18 Years JEE Main Chapter Wise Solved Disha

Table of contents :
Contents......Page 3
JEE Main Solved paper 2019 9 April......Page 9
JEE Main Solved paper 2019 9 January......Page 29
JEE Main Solved Paper - 2018......Page 57
1. Physical World, Units and Measurements......Page 85
2. Motion in a Straight line......Page 89
3. Motion in a Plane......Page 95
4. Laws of Motion......Page 99
5. Work, Energy and Power......Page 108
6. Rotational Motion......Page 116
7. Gravitation......Page 126
8. Mechanical Properties of Solids......Page 132
9. Mechanical Properties of Fluids......Page 134
10. Thermal Properties of Matter......Page 139
11. Thermodynamics......Page 145
12. Kinetic Theory......Page 152
13. Oscillations......Page 155
14. Waves......Page 165
15. Electric Charges and Fields......Page 172
16. Electrostatic Potential and Capacitance......Page 180
17. Current Electricity......Page 188
18. Moving Charges and Magnetism......Page 201
19. Magnetism and Matter......Page 211
20. Electromagnetic Induction......Page 214
21. Alternating Current......Page 217
22. Electromagnetic Waves......Page 224
23. Ray Optics and Optical Instruments......Page 226
24. Wave Optics......Page 234
25. Dual Nature of Radiation
and Matter......Page 240
26. Atoms......Page 246
27. Nuclei......Page 250
28. Semiconductor Electronics :
Materials, Devices and
Simple Circuits......Page 257
29. Communication Systems......Page 265
1. Some Basic Concepts of
Chemistry......Page 267
2. Structure of Atom......Page 272
3. Classification of Elements and
Periodicity in Properties......Page 278
4. Chemical Bonding and
Molecular Structure......Page 281
5. States of Matter......Page 288
6. Thermodynamics......Page 291
7. Equilibrium......Page 298
8. Redox Reactions......Page 307
9. Hydrogen......Page 309
10. The s-Block Elements......Page 311
11. The p-Block Elements
(Group-13 and 14)......Page 314
12. Organic Chemistry — Some
Basic Principles & Techniques......Page 317
13. Hydrocarbons......Page 327
14. Environmental Chemistry......Page 333
15. The Solid State......Page 334
16. Solutions......Page 337
17. Electrochemistry......Page 343
18. Chemical Kinetics......Page 351
19. Surface Chemistry......Page 358
20. General Principles and
Processes of Isolation of
Elements......Page 361
21. The p-Block Elements
(Group 15, 16, 17 & 18)......Page 363
22. The d-and f-Block Elements......Page 368
23. Co-ordination Compounds......Page 375
24. Haloalkanes and Haloarenes......Page 382
25. Alcohols, Phenols and
Ethers......Page 388
26. Aldehydes, Ketones and
Carboxylic Acids......Page 394
27. Amines......Page 399
28. Biomolecules......Page 404
29. Polymers......Page 409
30. Chemistry in Everyday Life......Page 412
31. Analytical Chemistry......Page 413
1. Sets......Page 415
2. Relations and Functions......Page 416
3. Trigonometric Functions......Page 417
4. Principle of Mathematical
Induction......Page 423
5. Complex Numbers and
Quadratic Equations......Page 424
6. Linear Inequality......Page 435
7. Permutations and
Combinations......Page 437
8. Binomial Theorem......Page 442
9. Sequence and Series......Page 448
10. Straight Lines &
Pair of Straight Lines......Page 458
11. Conic Sections......Page 469
12. Limits and Derivatives......Page 488
13. Mathematical Reasoning......Page 492
14. Statistics......Page 495
15. Probability......Page 500
16. Relations and Functions......Page 503
17. Inverse Trigonometric
Functions......Page 508
18. Matrices......Page 511
19. Determinants......Page 514
20. Continuity and Differentiability......Page 524
21. Application of Derivatives......Page 533
22. Integrals......Page 545
23. Applications of Integrals......Page 558
24. Differential Equations......Page 565
25. Vector Algebra......Page 571
26. Three Dimensional
Geometry......Page 590
27. Probability......Page 602
28. Properties of Triangles......Page 607

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EBD_7764

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JEE MAIN SOLVED PAPER–2018 PHYSICS

1.

2.

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is: (1) 2.5% (2) 3.5% (3) 4.5% (4) 6% All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. velocity

T m1 m1g

4.

(1)

distance time

(2)

5.

time

velocity

. Its total energy is:

k 4a

k

(2)

2

2a 2 -

(4)

n0 4

(2)

2n0

n0 n0 (4) 2 2 Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is: (3)

6.

3.

2r 2

3 k 2 a2 In a collinear collision, a particle with an initial speed n0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:

(1)

(4)

-

k

(3) zero

position

(3)

(1) 18.3 kg (2) 27.3 kg (3) 43.3 kg (4) 10.3 kg A particle is moving in a circular path of radius a under the action of an attractive potential

U= -

position

(1)

T

m m2 m 2

time

Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:

P O

EBD_7764 2018-2

(1)

JEE Main Solved Paper–2018

19 MR 2 2

(2)

55 MR 2 2

10.

73 181 MR 2 MR2 (4) 2 2 From a uniform circular disc of radius R and mass

(3) 7.

R is removed as 3 shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing, through centre of disc is :

9 M, a small disc of radius

11.

2R 3 R

(1) 4 MR2

(2)

40 MR2 9

37 MR2 9 A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then: (3) 10 MR2

8.

12.

(4)

13.

(1) T µ R3/2 for any n.(2) T µ R n /2+1 9.

(3) T µ R (n +1)/2 (4) T µ R n/2 A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional

æ dr ö decrement in the radius of the sphere çè ÷ø ,is : r (1)

Ka mg

(2)

Ka 3mg

(3)

mg 3Ka

(4)

mg Ka

14.

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy. (1) (a) 189 K (b) 2.7 kJ (2) (a)195 K (b) –2.7 kJ (3) (a)189 K (b) –2.7 kJ (4) (a)195 K (b) 2.7 kJ The mass of a hydrogen molecule is 3.32×10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly: (1) 2.35 × 103 N/m2 (2) 4.70 × 103 N/m2 (3) 2.35 × 102 N/m2 (4) 4.70 × 102 N/m2 A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10 12 /sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 ×1023 gm mole –1) (1) 6.4 N/m (2) 7.1 N/m (3) 2.2 N/m (4) 5.5 N/m A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations? (1) 5 kHz (2) 2.5 kHz (3) 10 kHz (4) 7.5 kHz Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is: é a 2 - b2 ù +cú ê ëê b ûú

(1)

s é a2 - b2 ù +c ú (2) ê Î0 ëê a ûú

s Î0

(3)

s é b2 - c2 ù +a ú (4) ê Î0 ëê b ûú

s é b 2 - c2 ù +a ú ê Î0 ëê c ûú

JEE Main Solved Paper–2018 15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric

16.

5 material of dielectric constant k = is inserted 3 between the plates, the magnitude of the induced charge will be: (1) 1.2 n C (2) 0.3 n C (3) 2.4 n C (4) 0.9 n C In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t

2018-3

(1) 2

(1) 50W, 10A (3) 17.

18.

19.

50 2

W, 0

(2)

21.

(1)

w0 L R

(2)

w0 R L

(3)

R (w0C)

(4)

CR w0

An EM wave from air enters a medium. The ur é æ z öù electric fields are E1 = E01 xµcos ê2pv ç - t ÷ ú in è c øû ë air and ur E 2 = E02 xµcos [ k(2z - ct)] in medium, where the

wave number k and frequency v refer to their

(4) 50W, 0

values in air. The medium is nonmagnetic. If Îr1

Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 W . The internal resistances of the two batteries

B The ratio 1 is: B2

1

the LC current exhibits resonance. The quality factor, Q is given by:

1000 W, 10A 2

are 1 W and 2 W respectively. The voltage across the load lies between: (1) 11.6 V and 11.7 V (2) 11.5 V and 11.6 V (3) 11.4 V and 11.5 V (4) 11.7 V and 11.8 V An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, ra respectively in a uniform magnetic field B. The relation between re, rp, ra is : (1) re > rp = ra (2) re < rp = ra (3) re < rp < ra (4) re < ra < rp The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2.

(4)

2

amplitude vm and frequency w0 =

pö æ i = 20 sin ç 30 t - ÷ 4ø è In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively:

3

1 2 For an RLC circuit driven with voltage of

(3) 20.

(2)

and Îr2 refer to relative permittivities of air and medium respectively, which of the following options is correct? Îr

(1)

1

Îr

=4

Îr

(2)

2

Îr

(3)

Îr

1

2

22.

=

1 4

1

Îr

Îr

(4)

Îr

=2

2 1 2

=

1 2

Unpolarized light of intensity I passes through an ideal polarizer A. Another indentical polarizer B is placed behind A. The intensity of light I . Now another 2 identical polarizer C is placed between A and B.

beyond B is found to be

The intensity beyond B is now found to be The angle between polarizer A and C is: (1) 0° (2) 30° (3) 45° (4) 60°

I . 8

EBD_7764 2018-4

23.

24.

JEE Main Solved Paper–2018

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (1) 25 m (2) 50 m (3) 75 m (4) 100 m An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n, g be the de Broglie wavelength of the electron in the n th state and

28.

29.

the ground state respectively. Let n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

25.

26.

n

A+

(3)

2 n

A+B

2 n 2 n

(2)

n

(4)

2 n

A+B

n

If the series limit frequency of the Lyman series is v1, then the series limit frequency of the Pfund series is : (1) 25 L (2) 16 L (3) L/16 (4) L/25 It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively: (1)

27.

B

(1)

89, 28

(2)

3V

CHEMISTRY 31.

28, 89

(3) (4) The reading of the ammeter for a silicon diode in the given circuit is :

200

30.

(1) 0 (2) 15 mA (3) mA (4) mA A telephonic communication service is working at carrier frequency of 10 GHz. Only10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz? (1) 2 × 103 (2) 2 × 104 5 (3) 2 × 10 (4) 2 × 106 In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 , a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell. (1) 1 (2) 1.5 (3) 2 (4) 2.5 On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1k . How much was the resistance on the left slot before interchanging the resistances? (1) (2) (3) (4)

32.

The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO 2 and H2O. The empirical formula of compound CXHYOZ is : (1) C3H6O3 (2) C2H4O (3) C3H4O2 (4) C2H4O3 Which type of ‘defect’ has the presence of cations in the interstitial sites? (1) Schottky defect (2) Vacancy defect (3) Frenkel defect (4) Metal deficiency defect

JEE Main Solved Paper–2018 33. According to molecular orbital theory, which of the following will not be a viable molecule? (1)

He22+

(2)

39.

He+2

(4) H 2– H2– 2 Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic reaction?

(3) 34.

ln K

A

B (0, 0) ++ ++ C ++ ++ D

35.

36.

37.

38.

40.

1 T(K)

(1) A and B (2) B and C (3) C and D (4) A and D The combustion of benzene (l) gives CO2 (g) and H2O (l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25°C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R= 8.314 JK–1 mol–1) (1) 4152.6 (2) –452.46 (3) 3260 (4) –3267.6 For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? (1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O (3) [Co(H2O)4Cl2]Cl. 2H2O (4) [Co(H2O)3Cl3].3H2O An aqueous solution contains 0.10 MH2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 × 10–7 and that of S2– from HS– ions is 1.2 × 10–13 then the concentration of S2– ions in aqueous solution is : (1) 5 × 10–8 (2) 3 × 10–20 (3) 6 × 10–21 (4) 5 × 10–19 An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is the original concentration of Ba2+ ? (1) 5 × 10–9 M (2) 2 × 10–9 M (3) 1.1 × 10–9 M (4) 1.0 × 10–10 M

41.

42.

43.

44.

45.

46.

2018-5 At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is : (1) 2 (2) 3 (3) 1 (4) 0 How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u) (1) 6.4 hours (2) 0.8 hours (3) 3.2 hours (4) 1.6 hours The recommended concentration of fluoride ion in drinking water is up to 1ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2. Ca(OH)2] to : (1) [CaF2] (2) [3(CaF2).Ca(OH)2] (3) [3Ca3(PO4)2.CaF2] (4) [3{(Ca(OH)2}.CaF2] Which of the following compounds contain(s) no covalent bond(s)? KCl, PH3, O2, B2H6, H2SO4 (1) KCl, B2H6, PH3 (2) KCl, H2SO4 (3) KCl (4) KCl, B2H6 Which of the following are Lewis acids? (1) PH3 and BCl3 (2) AlCl3 and SiCl4 (3) PH3 and SiCl4 (4) BCl3 and AlCl3

Total number of lone pair of electrons in I3– ion is : (1) 3 (2) 6 (3) 9 (4) 12 Which of the following salts is the most basic in aqueous solution? (1) Al(CN)3 (2) CH3COOK (3) FeCl3 (4) Pb(CH3COO)2 Hydrogen peroxide oxidises [Fe(CN)6]4– to [Fe(CN) 6]3– in acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium. The other products formed are respectively: (1) (H2O + O2) and H2O (2) (H2O + O2) and (H2O + OH–) (3) H2O and (H2O + O2) (4) H2O and (H2O + OH–)

EBD_7764 2018-6

47.

48.

49.

50.

51. 52.

53.

JEE Main Solved Paper–2018

The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O)2(NH3)] respectively are : (1) +3, +4, and +6 (2) +3, +2, and +4 (3) +3, 0, and +6 (4) +3, 0, and + 4 The compound that does not produce nitrogen gas by the thermal decomposition is : (1) Ba(N3)2 (2) (NH4)2Cr2O7 (3) NH4NO2 (4) (NH4)2SO4 When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is : (1) Zn (2) Ca (3) Al (4) Fe Consider the following reaction and statements: [Co(NH3)4Br2]+ + Br– ® [Co(NH3)3Br3] + NH3 (I) Two isomers are produced if the reactant complex ion is a cis-isomer. (II) Two isomers are produced if the reactant complex ion is a trans -isomer (III) Only one isomer is produced if the reactant complex ion is a trans -isomer (IV) Only one isomer is produced if the reactant complex ion is a cis-isomer. The correct statements are: (1) (I) and (II) (2) (I) and (III) (3) (III) and (IV) (4) (II) and (IV) Glucose on prolonged heating with HI gives : (1) n-Hexane (2) 1- Hexene (3) Hexanoic acid (4) 6-iodohexanal The trans-alkenes are formed by the reduction of alkynes with: (1) H2-Pd/C, BaSO4 (2) NaBH4 (3) Na/liq. NH3 (4) Sn - HCl Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation? NH2 (1) (2) –

N+2 Cl

NO2

(3)

54.

55.

56.

Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces :

(1)

(2)

(3)

(4)

An alkali is titrated against an acid with methyl orange as indicator, which of the following in a correct combination? Base Acid End point (1) Weak Strong Colourless to pink (2) Strong Strong Pinkish red to yellow (3) Weak Strong Yellow to Pinkish red (4) Strong Strong Pink to colourless The predominant form of histamine present in human blood is (pKa, Histidine – 6.0) (1)

H N N H N + N H

(2)

+

H N + N H

(3)

(4) (4)

H N

+ N

JEE Main Solved Paper–2018 2018-7 57. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br 2 to form product B. A and B are respectively :

(1)

(2)

and

(3)

and

(4)

58.

and

and

The increasing order of basicity of the following compounds is (2)

(a) (b)

(3)

(c)

59.

(d) (1) (a) < (b) < (c) < (d) (2) (b) < (a) < (c) < (d) (3) (b) < (a) < (d) < (c) (4) (d) < (b) < (a) < (c) The major product formed in the following reaction is :

(4) 60.

The major product of the following reaction is :

(1)

(2)

(3)

(4)

(1)

EBD_7764 2018-8

JEE Main Solved Paper–2018

MATHEMATICS 61.

66.

be the roots of the quadratic equation

The integral

18x 2 - 9 px + p 2 = 0 . Then the area (in sq.

sin 2 x cos 2 x

ò (sin5 x + cos3x sin 2 x + sin 3x cos2 x + cos5x)2 dx

units) bounded by the curve y = (gof)(x) and

is equal to :

the lines x = a, x = b and y = 0 , is :

(1)

(3)

62.

-1 3(1 + tan 3 x)

-1 1 + cot 3 x

+ C (2)

+C

(4)

1 1 + cot 3 x

+C

1 3(1 + tan3 x)

+C

(where C is a constant of integration) Tangents are drawn to the hyperbola

67.

4x 2 - y2 = 36 at the points P and Q. If these

63.

1 ( 3 + 1) 2

(2)

1 ( 3 - 2) 2

(3)

1 ( 2 - 1) 2

(4)

1 ( 3 - 1) 2

The sum of the co-efficients of all odd degree terms in the expansion of (x + x3 - 1)5 + (x - x3 - 1)5 ,(x > 1) is : (1) 0

(2) 1

(1)

(3) 2

(4) – 1

54 3

(2)

60 3

(3) 36 5 (4) 45 5 Tangent and normal are drawn at P(16, 16) on the

68.

(3)

4 3

(4)

1 2

12

k =0

2 a12 + a 22 + ... + a17 = 140m , then m is equal to :

(1) 68 (3) 33 9

r Let u be a vector coplanar with the vectors r r r a = 2iˆ + 3jˆ - kˆ and b = ˆj + kˆ . If u is perpenr r r r dicular to a and u × b - 24 , then | u |2 is equal

69.

to : (1) 315 (3) 84

70. (2) 256 (4) 336

If a, b Î C are the distinct roots, of the equation

Let a1,a 2 ,a 3 ,..., a 49 be in A.P. such that

å a 4k +1 = 416 and a 9 + a 43 = 66. If

the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ÐCPB = q , then a value of tan q is : (1) 2 (2) 3

65.

(1)

tangents intersect at the point T(0, 3) then the area (in sq. units) of DPTQ is :

parabola y2 = 16x , which intersect the axis of

64.

Let g(x) = cos x 2 ,f (x) = x , and a, b (a < b)

If

å (xi - 5) = 9 and i =1

(2) 34 (4) 66 9

å (x i - 5)2 = 45 , then the i=1

standard deviation of the 9 items x1, x2, ..., x9 is : (1) 4 (2) 2 (3) 3 (4) 9 PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45°, 30° and 30°, then the height of the tower (in m) is :

x 2 - x + 1 = 0 , then a101 + b107 is equal to :

(1) 50

(2)

(1) 0 (3) 2

(3)

(4) 100

(2) 1 (4) – 1

50 2

100 3

JEE Main Solved Paper–2018 71. Two sets A and B are as under :

76.

A = {(a, b) Î R ´ R :| a - 5 | < 1 and | b - 5 | < 1}; 2 2 B = {(a,b) Î R ´ R : 4(a - 6) + 9(b - 5) £ 36}. Then : (1) A Ì B (2) A Ç B = f (an empty set)

72.

73.

(3) neither A Ì B nor B Ì A (4) B Ì A From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is : (1) less than 500 (2) at least 500 but less than 750 (3) at least 750 but less than 1000 (4) at least 1000 Let

f (x) = x 2 +

1

and

x2

g(x) = x -

1 , x

f (x) x Î R - {-1,0,1} . If h(x) = , then the local g(x) minimum value of h(x) is : (1) – 3 (2) -2 2

(1)

77.

(1) (2) (3) (4)

75.

The value of

ò -

(1)

p 2

(3)

p 4

sin 2 x

p 1+ 2 2

x

dx is :

1 5

3 3 (4) 4 10 The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the

(1)

2 3

(2)

78.

1 3 2

2 (4) 3 3 If sum of all the solutions of the equation

(3)

æ æp ö æp ö 1ö 8cos x × ç cos ç + x ÷ × cos ç - x ÷ - ÷ - 1 è6 ø è6 ø 2ø è

(1)

13 9

(2)

8 9

20 2 (4) 9 3 A straight the through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :

(3) 79.

80.

(1)

2x + 3y = xy

(2)

3x + 2y = xy

(3)

3x + 2y = 6xy

(4)

3x + 2y = 6

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2 × 22 + 32 + 2.4 2 + 52 + 2.62 + ...

(2) 4p (4)

p 8

in

[0, p] is kp, then k is equal to :

is equal to 15. is equal to 120. does not exist (in R). is equal to 0. p 2

(2)

plane, x + y + z = 7 is:

(4) 3 2 2 For each t Î R , let [t] be the greatest integer less than or equal to t. Then æé 1 ù é 2 ù é15 ù ö lim x ç ê ú + ê ú + ... + ê ú ÷ è x x ë x ûø x ®0 + ë û ë û

2 5

(3)

(3)

74.

2018-9 A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

If B - 2A = 100 l , then l is equal to : (1) 248 (2) 464 (3) 496 (4) 232

EBD_7764 2018-10

81.

JEE Main Solved Paper–2018

If the curves y2 = 6x,9x 2 + by 2 = 16 intersect each other at right angles, then the value of b is : (1)

7 2

87.

2 | x - 3 | + x ( x - 6) + 6 = 0 . Then S : (1) (2) (3) (4)

(2) 4

9 (4) 6 2 Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is :

(3) 82.

83.

(1)

2 10

(3)

3 5 2

(2) (4)

85.

86.

5 2

88.

If

x - 4 2x

2x

2x

89.

10

Let y - y(x) be the solution of the differential equation sin x

x-4

90.

(1)

-8 2 p 9 3

(2)

(3)

4 - p2 9

(4)

has a non-zero solution (x, y, z), then (2) – 30 (4) – 10

y2

8 - p2 9

4 9 3

p2

If L1 is the line of intersection of the planes

2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is the line of intersection of the planes

~ (pÚ q) Ú (~ p Ù q) is equivalent to : (1) p (2) q (3) ~q (4) ~p If the system of linear equations x + ky + 3z = 0 3x + ky – 2z = 0 2x + 4y – 3z = 0 xz

dy + y cos x = 4x, x Î (0, p) . If dx

æ pö æ pö y ç ÷ = 0 , then y ç ÷ is equal to : è2ø è6ø

= (A+ Bx)(x - A) 2 ,

then the ordered pair (A, B) is equal to : (1) (– 4, 3) (2) (– 4, 5) (3) (4, 5) (4) (– 4, – 5) The Boolean expression

to : (1) 10 (3) 30

If the tangent at (1, 7) to the curve x 2 = y - 6

then the value of c is : (1) 185 (2) 85 (3) 95 (4) 195

2x

2x

contains exactly one element. contains exactly two elements. contains exactly four elements. is an empty set.

touches the circle x 2 + y2 + 16x + 12y + c = 0

Let S = { t Î R : f (x) =| x - p | (e|x| - 1)sin | x | is not differentiable at t}. Then the set S is equal to : (1) {0} (2) {p} (3) {0, p} (4) f (an empty set) x - 4 2x

84.

3

Let S = {x Î R : x ³ 0 and

x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0 , then the distance of the origin from the plane, containing the lines L1 and L2, is : (1)

is equal

(3)

1 3 2 1 2

(2)

(4)

1 2 2 1 4 2

JEE Main Solved Paper–2018

2018-11

Hints and Solutions PHYSICS 1.

(3)

Density (d) =

Mass (M) M = Volume (V) L3

Dd DM 3DL = + d M L = 1.5% + 3(1%) = 4.5% (2) Graphs in option (3) position-time and option (1) velocity-position ar e corresponding to velocity-time graph option (4) and its distance-time graph is as given below. Hence distance-time graph option (2) is incorrect.

\ Error in density,

2.

100 = m + 10 \ m = 23.3kg ; close to 3 option (2)

4.

(3)

¶u K rˆ = 3 rˆ ¶r r Since particle is moving in circular path F=-

F=

mv 2 K K = Þ mv 2 = 3 r r r2

1 K \ K.E. = mv2 = 2 2 2r Total energy = P.E. + K.E.

distance

=-

K 2r

2

+

K 2r 2

= Zero (Q P.E. = -

time

5.

(2) Given : m1 = 5kg; m2 = 10kg; m = 0.15 FBD for m1, m1g – T = m1a Þ 50 – T = 5 × a and, T – 0.15 (m + 10) g = (10 + m)a For rest a = 0 or, 50 = 0.15 (m + 10) 10

N m T m2 m(m+m2)g (m+m2)g

T m1 m1g = 50N

Þ

5=

3 (m + 10) 20

2r 2

given)

(2) Before Collision

3

K

m

V0

After Collision m

Þ

m

V1

V2

m

Stationary

1 2 1 2 3æ1 ö mv + mv = ç mv02 ÷ 2 1 2 2 2è2 ø 3 Þ v12 + v 22 = v20 2 From momentum conservation mv0 = m(v1 + v2) Squarring both sides, (v1 + v2)2 = v02

....(i) ....(ii)

Þ v12 + v22 + 2v1v2 = v02 2v1v2 = -

v20 2

(v1 - v 2 ) 2 = v 21 + v 22 - 2v1v2 =

v2 3 2 v0 + 0 2 2

EBD_7764 2018-12

JEE Main Solved Paper–2018

Moment of inertia (I1) of the complete disc

Solving we get relative velocity between the two particles v1 - v2 = 2v0 6.

1 9 MR 2 about an axis passing through 2 O and perpendicular to the plane of the disc. M.I. of the cut out portion about an axis passing through O' and perpendicular to the plane of disc

=

(4) Using parallel axes theorem, moment of inertia about ‘O’ Io = Icm + md2

=

7MR 2 55MR 2 + 6(M ´ (2R)2 ) = 2 2

=

2R O

2 2 é1 æ Rö æ 2R ö ù = ê ´M ´ç ÷ +M ´ç ÷ ú è 3ø è 3 ø ú êë 2 û [Using perpendicular axis theorem] \ The total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is I = I1 + I2 =

2R 2R 2R

Again, moment of inertia about point P, Ip = Io + md 2

55MR 2 181 + 7M(3R) 2 = MR 2 2 2 (1) Let s be the mass per unit area. =

7.

2 2 é1 1 æ Rö æ 2R ö ù 9 MR 2 - ê ´ M ´ ç ÷ + M ´ ç ÷ ú è 3ø è 3 ø ú 2 êë 2 û

R/ 3 O'

= 2R/3 O

8.

(3)

9MR 2 9MR 2 (9 - 1)MR 2 = 4 MR 2 = 2 18 2

mw2 R = Force µ

(Force = The total mass of the disc = s × p R2 = 9M The mass of the circular disc cut

æ Rö = s´pç ÷ è 3ø

2

p R2 =M = s´ 9

Let us consider the above system as a complete disc of mass 9M and a negative mass M super imposed on it.

2

\ M.I. (I2) of the cut out portion about an axis passing thr ough O and perpendicular to the plane of disc

2R

2R

1 æ Rö ´M ´ç ÷ è 3ø 2

1 Rn

mv2 ) R

Þ w2 µ

1

Þ wµ

R n +1

Time period T =

2p w

Time period, T µ R

n +1 2

1 n +1 R 2

JEE Main Solved Paper–2018 9.

(3)

Bulk modulus, K = K=

2018-13

volumetric stress volumetricstrain

mg æ dV ö aç ÷ è V ø

dV mg Þ = V Ka

dV 3dr = V r ...(ii)

3dr mg = r Ka

12.

(

5 3

300

13.

f Change in internal energy DU = n R DT 2

æ 3 öæ 25 ö = 2 ç ÷ç ÷ (-111) = -2.7 kJ è 2 øè 3 ø (1) Change in momentum

P ^J 2

– P J^ 2

108

)

´ 10-3 kg

1 k ´ 6.02 ´ 1023 = 1012 2p 108 ´ 10-3 Solving we get, spring constant, K = 7.1N/m (1) In solids, Velocity of wave V=

Y = r

9.27 ´ 1010

2.7 ´ 103 v = 5.85 × 103 m/sec Since rod is clamped at middle fundamental wave shape is as follow l = L Þ l = 2L 2

j^

45°

2 ´ 10 -4 =2.35 × 103N/m2 (2) As we know, frequency in SHM

1 k = 1012 2p m where m = mass of one atom Mass of one atom of silver,

(300)V2/3 = T2(2V)2/3

11.

2 ´ 3.32 ´ 10 -27 ´ 103 ´ 1023

= 6.02 ´1023

(2) 2/3 T2 = 189 K (final temperature)

(Q n = no.of particles)

f=

or, T1V1g–1 = T2V2g–1

Þ T2 =

2P ˆ J 2

F 2P = n A A

=

dr mg = (fractionaldecrement in radius) r 3Ka (3) In an adiabatic process TVg–1 = Constant

For monoatomic gas g =

2P ˆ J = IH molecule 2

=

\

10.

DP =

Pressure, P

4 3 volume of sphere , V = pR 3

Using eq. (i) & (ii)

P ˆ P ˆ P ˆ P ˆ J+ J+ ii 2 2 2 2

Þ Iwall = -

....(i)

Fractional change in volume

DP =

P P^ 45° 2 i P ^i 2 P

A

i^

A N l/2 l = 1.2m (Q L = 60 cm = 0.6m (given) Using v = fl

EBD_7764 2018-14

JEE Main Solved Paper–2018

v 5.85 ´103 = l 1.2 = 4.88 × 103 Hz ; 5 KHz (2) Potential outside the shell, Þ

14.

Pavg =

f=

KQ r where r is distance of point from the centre of shell

=

17.

KQ R

2

watt

sin q =

12V v–12

S

20 2

1W

U

2W

P Q 0

10W

Using Kirchhoff’s law at P we get V - 12 V - 13 V - 0 + + =0 1 2 10 [Let potential at P, Q, U = 0 and at R = V

b c

Þ

V V V 12 13 0 + + = + + 1 2 10 1 2 10 10 + 5 + 1 24 + 13 V= 10 2

VB =

Kq A Kq B KqC + + rb rb rc

Þ

VB =

1 é s4pa 2 s4pb 2 s4pc 2 ù + ê ú 4p Î0 ëê b b c ûú

æ 16 ö 37 Þ Vç ÷ = è 10 ø 2

é a 2 - b2 ù + cú ê êë b úû (1) Charge on Capacitor, Qi = CV After inserting dielectric of dielectric constant = K Qf = (kC) V Induced charges on dielectric VB =

s Î0

sin 45° = 10A

v–13

R V

s C –s B s a A

16.

I0

(2) T

where ‘R” is radius of the shell

15.

2

Wattless current I = Irms sin q

Voutside =

Potential inside the shell, Vinside =

1000

37 ´10 370 = = 11.56 volt 16 ´ 2 32 (2) As we know, radius of circular path in magnetic field ÞV=

18.

r=

2Km qB

Qind = Qf - Qi = KCV - CV æ5 ö = (K - 1)CV = ç - 1 ÷ ´ 90pF ´ 2V = 1.2nc è3 ø

For electron, re =

(2) As we know, average power Pavg = Vrms Irms cosq

For proton, rp =

2Km e

....(i)

eB

2Km p

....(ii)

eB

For a particle,

æ V öæ I ö æ 100 ö æ 20 ö = ç 0 ÷ ç 0 ÷ cos q = ç ÷ç ÷ cos 45° 2 2 è øè ø è 2 øè 2 ø (Q q = 45°)

ra =

\

2Km a qa B

r e < rp = ra

=

2K4m p 2eB

=

(Q me < mp)

2Km p eB

...(iii)

JEE Main Solved Paper–2018 19.

2018-15 After introducing polariser C between A and B,

(3) Magnetic field at the centre of loop, m0 I 2R Dipole moment of circular loop is m = IA m1 = I.A = I.pR2 {R = Radius of the loop} If moment is doubled (keeping current constant) B1 =

R becomes

2R

(

)

m 2 = I.p

B2 =

\

2

(

2R

m0 I 2R

B1 = B2 2

(

2

2R

23. = 2

) w0 w L = 0 2Dw R

(3) Velocity of EM wave is given by

1

v=

mÎ

w Velocity in air = k = C

22.

C =2 æCö ç ÷ è2ø

Þ

Îr1 Îr2

=

1 4

(3) Axis of transmission of A & B are parallel.

Polariser A

24.

Polariser B

I/2

2

or, q = 45° 2l d

l´D d ; Fringe width, b = d' 2

d 50 ´ 10-2 10 -6 ´ 50 ´10 -2 10-2 = ´ = 2 d' 2´d' Therefore, slit separation distance, d’ = 25mm (1) Wavelength of emitted photon from nth state to the ground state,

1 æ 1 ö 1÷ 2 ç RZ è n 2 ø Since n is very large, using binomial theorem Ln =

Ln = Ln =

1 æ 1 ö 1+ ÷ 2 ç RZ è n 2 ø 1 2

+

RZ As we know,

ln =

I

1

-1

Here, m1 = m2 = 1 as medium is non-magnetic

=

I/2Cos4 q

1 1 ö æ1 = RZ 2 ç ÷ 2 Ln è1 n2 ø

C Velocity in medium = 2

1 Îr2

2

(1) Angular width of central maxima = or, l =

21.

Polariser B

I/2Cos q

I/2

Þ cos q =

)

m0 I 2R m0 I

Polariser C

I I 1 cos4 q = Þ cos4 q = 2 8 4

= 2.IpR = 2m1

(1) Quality factor Q =

\

I

2

20.

1 Îr1

Polariser A

1 æ 1 ö ç ÷ RZ 2 è n 2 ø

æ n 2h2 ö 1 2pr = 2p ç ÷ µn ç 4p2 mZe2 ÷ n n è ø

I/2 Ln » A +

B ln2

EBD_7764 2018-16

25.

(4)

JEE Main Solved Paper–2018

hnL = E ¥ - E1

...(i)

hnf = E ¥ - E5

...(ii)

Eµ

z2 n

2

2

Þ

E5 æ 1 ö 1 =ç ÷ = E1 è 5 ø 25

Eqn (i) / (ii) Þ Þ

26.

hn L E1 = hn f E 5

29.

v 3

1 1 mv2 - mv12 8 2 2 = = 0.89 Pd = 1 9 2 mv 2

Now, For collision of neutron with carbon nucleus v v1 v2 m m 12m 12m Applying Conservation of momentum mv + 0 = mv1+ 12mv2 ....(iii) v = v2 – v1 ....(iv) From eqn (iii) and eqn (iv) 11 v 13 2

1 1 æ 11 ö mv2 - m ç v ÷ 48 2 2 è 13 ø = » 0.28 Pc = 1 169 mv2 2

27.

(3) Clearly from fig. given in question, Silicon diode is in forward bias. \ Potential barrier across diode DV = 0.7 volts

10 ´ 10 ´ 109 = n ´ 5 ´ 103 100 Þ n = 2 × 105 (2) Using formula, internal resistance,

or,

(1) For collision of neutron with deuterium: v v1 v2 m m 2m 2m Applying conservation of momentum : mv + 0 = mv1 + 2mv2 .....(i) v2 – v1 = v .....(ii) Q Collision is elastic, e = 1 From eqn (i) and eqn (ii) v1 = -

V - DV 3 - 0.7 2.3 = = = 11.5mA R 200 200 (3) If n = no. of channels 10% of 10 GHz = n × 5 KHz I=

28.

n L 25 n = Þ nf = L nf 1 25

v1 = -

Current,

æ l -l r =ç 1 2 è l2

30.

ö ÷s ø

æ 52 - 40 ö =ç ÷ ´ 5 = 1.5W è 40 ø (3) R1 + R2 = 1000 Þ R2 = 1000 – R1

R1

R2 = 1000 – R1 G

(l ) 100 – l On balancing condition R1(100 – l) = (1000 – R1)l ...(i) On Interchanging resistance balance point shifts left by 10 cm R2=1000 – R1

R1 G

(l – 10)

(100 – l + 10) =(110 – l)

On balancing condition (1000 – R1) (110 – l) = R1 (l – 10) or, R1 (l – 10) = (1000 – R1) (110 – l) ...(ii) Dividing eqn (i) by (ii) 100 - l l = l - 10 110 - l Þ (100 – l) (110 – l) = l(l – 10) Þ 11000 – 100l – 110l + l2 = l2 – 10l Þ 11000 = 200l or, l = 55

JEE Main Solved Paper–2018 Putting the value of ‘l’ in eqn (i) R1 (100 – 55) = (1000 – R1) 55 Þ R1 (45) = (1000 – R1) 55 Þ R1 (9) = (1000 – R1) 11 Þ 20 R1 = 11000 \ R1 = 550KW

2018-17

CHEMISTRY 31.

(4)

34.

Element C

6

H

1

6 = 0.5 12 1 =1 1

s* 1s 2 1s 2

s

s

1s 2

2-2 =0 2 2-0 =1 2

Molecule having zero bond order will not be a viable molecule. (1) From thermodynamics -DH° DS° + RT R

for exothermic reaction, DH = –ve

2

slope =

– DH° = + ve R

So from graph, line should be A & B. 35.

(4) C 6 H 6 ( l) +

15 O 2 ( g ) ¾¾ ® 6CO 2 ( g ) + 3H 2 O ( l ) 2

Dn g = 6 -

Yö æ 2 ç X + ÷ = 2Z è 4ø

15 3 =2 2

DH = DU + Dn g RT æ 3ö = -3263.9 + ç - ÷ ´ 8.314 ´ 298 ´ 10-3 è 2ø

æ 2ö çè1 + ÷ø = Z Þ Z = 1.5 4 \ molecule can be written as CXHYOZ Þ

C1H2O3/2 Þ C2H4O3 (3) In Frenkel defect some of ion (usually cation due to their small size) missing from their normal position and occupies position in interstitial position. (4) Electronic configuration Bond order

H-2

(Z = 2)

1

Yö æ Oxygen atoms required = 2 çè X + ÷ø 4 As mentioned,

He +2

He 22+

ln K =

Yö Y æ CX H Y + ç X + ÷ O2 ¾¾ ® XCO 2 + H 2O è 4ø 2

33.

(Z = 4)

Relative Relative Simplest whole mass mole number ratio

So, X = 1, Y = 2 Equation for combustion of CXHY

32.

H 22 -

(Z = 3)

s 2 s* 1 1s 1s

2 -1 = 0.5 2

(Z = 3)

s

s* 1s 2 1s1

2 -1 = 0.5 2

= – 3263.9 + (–3.71) = – 3267.6 kJ mol–1 36.

(4) Number of particles (i) (1) [Co(H2O)6]Cl3

1

(2) [Co(H2O)5Cl]Cl2.H2O

3

(3) [Co(H2O)4Cl2]Cl.2H2O

2

(4) [Co(H2O)3Cl3].3H2O

4

DTf µ i ; where DTf = (Tf - Tf¢ ) Remember, the greater the no. of particles, the lower will be the freezing point. Compound (d) will have the highest freezing point due to least no of particles.

EBD_7764 2018-18

37.

JEE Main Solved Paper–2018

(2) In presence of external

H+,

r1 = 1 torr s–1, when 5% reacted r2 = 0.5 torr s–1, when 33 % reacted

ˆˆ† 2H + + S2 - , K a . K a = K eq H 2S ‡ˆˆ 1 2

(a - x1 ) = 0.95(unreacted) (a - x 2 ) = 0.67(unreacted)

2

\

éH + ù éS2- ù ë û ë û = 1 ´ 10-7 ´ 1.2 ´ 10-13 [ H2S]

m

1 r1 é (a - x1 ) ù æ 0.95 ö =ê =ç ú ; ÷ r2 ë (a - x 2 ) û 0.5 è 0.67 ø

[ 0.2]2 éëS2- ùû = 1.2 ´ 10 -20 0.1 [ ]

m

2 = (1.41)m Þ 2 = ( 2)m Þm=2

2- ù

éS ë 38.

û

= 3 ´ 10

-20

40.

1M + Final Ba2+ ¾® Solution Na2SO 4 50 mL 450 mL 500 mL

On electrolysis : 6H 2 O ¾¾ ® 6H 2 + 3O 2

Concentration of SO 2– 4 in BaSO4 solution

Number of Faradays = 12 = Amount of charge 12 × 96500 = i × t 12 × 96500 = 100 × t

M1V1 = M2V2 1 × 50 = M2 × 500 1 10

t=

For just precipitation Ionic product = Ksp

=

[Ba2+] [SO42–] = Ksp(BaSO4) [Ba2+] ×

1 = 10–10 10

[Ba2+] = 10–9 M in 500 mL solution

41.

(1)

Generally r µ (a – x)m

(3) [3Ca 3 (PO 4 )2 .Ca(OH) 2 ]+

2F –

(drinking water upto1ppm)

¾¾ ®

[3Ca 3(PO 4 ) 2 .CaF2 ] + 2OH – Fluorapatite (Harder teeth enamel)

500 ´ 10 –9 = 1.11 ´ 10-9 M 450

CH 3CHO ¾¾ ® CH 4 + CO

12 ´ 96500 hour 100 ´ 3600

Hydroxyapatite

Þ M1 × 450 = 10–9 × 500 [ where M1 = molarity of original solution]

39.

12 ´ 96500 second 100

= 3.2 hours

Thus [Ba2+] in original solution (450 mL)

M1 =

B2 H 6 + 3O 2 ¾¾ ® B2 O 3 + 3H 2 O

27.66 g of B2H6 (1 mole) requires 3 moles of oxygen (O2) for complete burning. Required O2 (3 moles) is obtained by electrolysis of 6 moles of H2O

(3)

M2 =

(3)

42.

(3) KCl is an ionic compound while other (PH3, O2, B2H6, H2SO4) are covalent compounds.

JEE Main Solved Paper–2018 43.

(2, 4) BCl3 and AlCl3, both have vacant p-orbital and incomplete octet thus they behave as Lewis acids. SiCl4 can accept lone pair of electron in dorbital of silicon hence it can act as Lewis acid. Although the most suitable answer is (c). However, both options (c) and (a) can be considered as correct answers. e.g. hydrolysis of SiCl4 Cl

Cl

Si Cl

Cl + Cl

H2O

¾®

Cl

Si Cl Cl

47.

48.

H O

H

Cl Cl — Si — OH + HCl

49.

Cl

(3)

D

(1)

Ba(N 3 ) 2 ¾¾® Ba + 3N 2

(2)

(NH 4 )2 Cr2 O7 ¾¾® Cr2 O3 + N 2 + 4H 2O

(3)

NH 4 NO 2 ¾¾® N 2 + 2H 2 O

(4)

(NH 4 ) 2 SO 4 ¾¾® 2NH 3 + H 2 SO 4

(3)

D

D

D

NH3 is evolved in reaction (c). NaOH

Al + 3H 2 O ¾¾¾¾ ® Al(OH)3 ¯ + 3 / 2H 2 (g)

i.e., option (a) AlCl 3 and SiCl 4 is also correct. 44.

2018-19 (3) [Cr (H2O)6] Cl3 Þ x + 6 × 0 + (– 1) × 3 = 0 Þ x = +3 [Cr (C6H6)2] x+2×0= 0;x=0 K2 [Cr (CN)2 (O)2 (O2) (NH3)] 2 × 1 + x + 2 × (– 1) + 2 × (– 2) + (– 2) + 0 = 0 x=+6 (4)

(X)

White gelatinous ppt.

I excess of NaOH

Al(OH)3 ¾¾¾¾¾¾® Na[Al(OH) 4 ] Soluble

I D

® Al2O3 + 3H 2O 2Al(OH)3 ¾¾

I 45.

46.

\ Total number of lone pair of electrons is 9. (2) CH 3 COOK is a salt of weak acid (CH3COOH) and strong base (KOH). FeCl3 is a salt of weak base [Fe(OH)3] and strong acid (HCl). Pb (CH3COO)2 is a salt of weak base Pb(OH)2 and weak acid (CH3COOH) Al(CN)3 is a salt of weak base [Al (OH)3] and weak acid (HCN). 1 (3) [Fe(CN)6]4– + H2O2 + H+ ¾® [Fe(CN)6]3– 2 + H2O 1 – H O + OH 2 2 2 1 ¾® [Fe(CN)6]4– + H2O + O2 2

[Fe(CN)6]3– +

(X )

Al2O3 is used as adsorbent in chromatography. Thus, metal ‘M’ is Al. 50.

(2) H3N

H3N

Br Br

NH3 cis-isomer +Br –

NH3

EBD_7764 2018-20

JEE Main Solved Paper–2018

Br H3N

H3N

Br

Br Br

+ H3N

Br

H3N

NH3

Br

NH3

mer–

fac– (2 isomers)

H3N

Br

Br H3N

NH3

NH3

Br –

¾+ ¾ ® H3N

NH3

H3N

Br

Br Br mer (1 isomer)

trans

51.

55.

(1)

(3) pH range for methyl orange is ¬¾¾¾¾¾ 3.9 - 4.5 ¾¾¾¾ ®

CHO (CH — OH)4

Pinkish red

HI, D

¾® CH3CH2CH2CH2CH2CH3 n-Hexane

CH2 — OH

Na /liq.NH

3 ¾¾¾¾¾¾ ® Birch reduction

52.

(3)

53.

(1) Kjeldahl’s method is not applicable for compounds containing nitrogen in nitro and azo groups and nitrogen in ring, as N of these compounds does not change to ammonium sulphate under these conditions. (1)

54.

OH

56.

Generally, weak bases have pH greater than 7. When methyl orange is added to a weak base solution, solution becomes yellow. This solution is then titrated by a strong acid and at the end point pH will be less than 3.1. \ Solution becomes pinkish red. (4) Structure of histamine H

N

OH COOH

CO2, NaOH

¾¾¾¾¾® Acidification

\

(X) (Major)

O

(CH CO) O

O—C—CH3 COOH

2 ¾ ¾3 ¾ ¾ ®

H2SO4

Acetyl salicylic acid (Aspirin)

Yellow

more basic

N Blood is slightly basic in nature (7.35 pH). At this pH, terminal NH2 will get protonated due to more basic nature. Predominant structure of histamine is H N N

JEE Main Solved Paper–2018

2018-21

O –

OH

57.

(3)

O

O — C — O — CH3

O Cl — C — O — CH3

–

OH ¾ ®

¾¾¾¾¾¾®

(A)

O O — C — O — CH3 Br

2 ¾¾¾¾ ¾®

Br (B)

58.

(3) (a) (1° and

(b)

+

NH2

(sp2 )

sp3

Protonation

¾¾¾¾® )

+

NH2

NH Protonation ¾¾¾¾ ®

NH2

(c)

NH3

+

NH2 Protonation

¾¾¾¾® NH

NH2 ¾¾® + NH2

NH2

[Equivalent resonance]

(d) (2° and sp3)

NHCH3

¾Protonation ¾¾¾®

+

NH2

CH3

Hence, correct order of basicity will be : b < a < d < c. 59.

(4) +

¾ ¾®

(–C2H 5I)

– CH 3OH +

EBD_7764 2018-22

60.

JEE Main Solved Paper–2018

O–

(2) CH 3 is a strong base and strong nucleophile, so favourable condition is SN2/E2. The given alkyl halide is 2° and b carbons are 4° and 2°, so sufficiently hindered, thus E2 dominates over SN2. Also, polarity of CH3OH (solvent) is not as high as H2O, so E1 is also dominated by E2.

\

Þ y = –12

Y

4° b

a

X'

–

CH O E2

3 ¾¾®

H

(Major product)

X

Q

P

R

(2°)

Y'

MATHEMATICS 61.

(1) Let I =ò

sin2 xcos2 x (sin5x+ cos3xsin2x+ sin3xcos2 x + cos5x)2

=ò =ò

=ò

2

dx

2

sin x cos x 2

2

3

3

2

[(sin x + cos x)(sin x + cos x)] sin 2 x cos 2 x (sin 3 x + cos3 x)2 tan 2 x × sec 2 x (1 + tan 3x)2

dx

dx

\

1 Area of DPQT = ´ TR ´ PQ 2

Q

P º (3 5, -12) \ TR = 3 + 12 = 15,

\

Area of

1 DPQT = ´15 ´ 6 5 = 45 5 sq. units 2 63.

(1) Equation of tnagent at P(16, 16) is given as: x – 2y + 16 = 0

dx

Y

tan3x) = t

\

I=

-1 1 dt 1 +C =- + C = 3 ò t2 3t 3(1 + tan3 x)

X'

t en ng a T A (–16, 0)

, 16 P(

) 16

al rm No

Now, put (1 + Þ 3 tan2x sec2x dx = dt

62.

x 2 y2 =1 9 36

(0, 3)

Br b

x(0) y(3) =1 9 36

Equation of PQ is :

B(24, 0)

(4) Here equation of hyperbola is

x2 y2 =1 9 36 Now, PQ is the chord of contant

Y' Slope of PC (m1) =

4 3

X

JEE Main Solved Paper–2018

2018-23

Slope of PB (m2) = –2

12

4 +2 m1 - m 2 Hence, tan q = = 3 1 + m1 × m 2 1 - 4 × 2 3

64.

68.

13 [2a1 + 48d] = 416 2 Þ a1 + 24d = 32 ...(1) Now, a9 + a43 = 66 Þ 2a1 + 50d = 66 ...(2) From eq. (1) & (2) we get; d = 1 and a1 = 8

r r r r r = l{a 2 .b - (a .b)a}

Also,

ˆ = l'{- iˆ+ 2 ˆj+ 4k}. ˆ = l{-4iˆ + 8jˆ + 16k} rr Also, u.b = 24 Þ l' = 4 \ Þ 65.

66.

r2 u = 336

å (r + 7)2 = 140 m

Þ

å (r 2 + 14r + 49) = 140 m

69.

(x +

Also,

Þ

ò

) + (x -

x3 - 1

)

9

i =1

i =1

å (x i - 5)2 = 45 i=1

9

9

i =1

i =1

å x i2 - 10å xi + 9(25) = 45 9

å xi2 = 360 i =1

Since, variance =

5

= 2[5C0 x5 + 5C2 x3(x3 – 1) + 5C4x(x3 – 1)2] Þ 2[x5 + 10x6 – 10x3 + 5x7 – 10x4 + 5x] \ Sum of coefficients of odd degree terms = 2.

...(ii)

From (i) and (ii) we get,

cos xdx = 3 - 1 2 p /6

x3 - 1

9

å (x i - 5) = 9 Þ å xi = 54 ...(i) 9

(3) Since we know that, (x + a)5 + (x – a)5 = 2[5C0 x5 + 5C2 x3× a2 + 5C4x×a4] \

r =1

(2) Given

p p a= ,b= 6 3

5

r =1

æ 17 ´18 ´ 35 ö æ 17 ´18 ö ç ÷ + 14 ç ÷ + (49 ´17) = 140 è ø è 2 ø 6 Þ m = 34

p /3

67.

r =1

Þ

a = -w and b = -w2

Req. area =

r =1

17

Also, gof(x) = cosx \

17

Þ

r u = -4iˆ + 8jˆ + 16kˆ

where w is cube root of unity \ a101 + b107 = (–w)101 + (–w)107 = –[w2 + w] = –[–1] = 1 (4) Here, 18x2 – 9px + p2 = 0 Þ (3x – p) (6x – p) = 0 Þ

17

å a 2r = å [8 + (r - 1)1]2 = 140 m

17

(2) a, b are roots of x2 – x + 1 = 0 \

k =0

Þ

Þ tan q = 2 r r r (4) Q u, a & b are coplanar

r r r r \ u = l(a´ b) ´ a

(2) Q å a 4k +1 = 416

å xi 2 - æ å x i ö 9

2

ç 9 ÷ è ø

2

= \

360 æ 54 ö - ç ÷ = 40 – 36 = 4 9 è9ø

Standard deviation = Variance = 2

EBD_7764 2018-24

70.

JEE Main Solved Paper–2018

(4)

72.

P

6

73. 90°

h

30°

30° M Let height of tower MN = h In DQMN we have

tan 30° = \

\

\ ...(1) ...(2)

\

(1) A ={(a,b) Î R ´ R :| a - 5 |< 1,| b - 5 |< 1} Let a – 5, = x, b – 5 = y Set A contains all points inside 2 2 B = {(a,b) ÎR ´ R: 4(a - 6) + 9(B- 5) £ 36} Set B contains all points inside or on

(x - 1)2 y 2 + =1 9 4 Y

\

(1, 0) (1, –1) (–1, –1) (0, –1)

Y¢ (±1, ±1) lies inside the ellipse.

Hence, A Ì B .

1 x

£ -2 2

x-

1 >0 x

1 2 + ³2 2 x x- 1 x

74.

æ 1 2 15 ö (2) Since, lim x ç é ù + é ù + ... + é ù ÷ ê ú ê ú êë x úû ø x ® 0+ è ë x û ë x û

æ 1 + 2 + 3 + ... + 15 ö = lim+ x ç ÷ø è x x®0 æì 1 ü ì 2 ü ì15 ü ö çè í x ý + í x ý + ... + í x ý ÷ø î þ î þ î þ

(0, 1)

(–1, 0)

x-

Hence, 2 2 will be local minimum value of h(x).

| x |< 1,| y |< 1

X¢

2

When x -

From (2), we get :

(0, 1)

1 + x

1 vB (d) their velocities depend on their masses. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is [2003] (a) 12 m (b) 18 m (c) 24 m (d) 6 m A ball is released from the top of a tower of height h meters. It takes T seconds to reach the T ground. What is the position of the ball at 3 second [2004] (a)

8h meters from the ground 9

7h meters from the ground 9 h (c) meters from the ground 9 17 h meters from the ground (d) 18 An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be [2004] (a) 60 m (b) 40 m (c) 20 m (d) 80 m A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate

(b)

6.

7.

8.

f to come to rest. If the total distance traversed 2 is 15 S , then [2005] 1 (a) S = ft 2 (b) S = f t 6 1 2 1 2 ft (c) S = ft (d) S = 4 72 A particle is moving eastwards with a velocity of 5 ms–1. In 10 seconds the velocity changes to 5 ms–1 northwards. The average acceleration in this time is [2005] 1 -2 ms towards north 2 1 (b) ms - 2 towards north-east 2 1 (c) ms - 2 towards north-west 2 (d) zero

(a)

EBD_7764 P-6

9.

10.

11.

12.

Physics

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is [2005] 3 2 (a) –2bv (b) –2abv (c) 2av2 (d) –2av 3 A particle located at x = 0 at time t = 0, starts moving along with the positive x-direction with a velocity 'v' that varies as v = a x . The displacement of the particle varies with time as [2006] (a) t 2 (b) t (c) t 1/2 (d) t 3 The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is [2007] (a) v0 + g /2 + f (b) v0 + 2g + 3f (c) v0 + g /2 + f/3 (d) v0 + g + f A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x1(t) after time ‘t’; and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time ‘t’? [2008]

(x1 – x2)

(d)

t

O

13. Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be : [2009] v +v1 t

O

(a)

–v1 y h

t

v +v1 (b) O

t1

2t1

t

4t1

–v1

(x1 – x2)

y h

(a)

O

t

t

t

(x1 – x2)

(b)

O

t

(x1 – x2)

(c)

O

t1

2t1

t

y h

(c)

O

t

t

P-7

Motion in a Straight Line

15. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is: [2014] (a) 2gH = n2u2 (b) gH = (n – 2)2 u2d (c) 2gH = nu2 (n – 2) (d) gH = (n – 2)u2 16. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? [2017]

v v1 (d)

t

O y h

t 14.

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by

(a)

(b)

(c)

(d)

dv = -2.5 v where v is the instantaneous dt

speed. The time taken by the object, to come to rest, would be: [2011] (a) 2 s (b) 4 s (c) 8 s (d) 1 s

Answer Key 1 (a) 16 (a)

1.

2 (d)

3 (b)

4 (c)

5 (a)

6 (d)

7 (d)

(a) Activity A to B u u1 = u ; v1 = , s1 = 0.03 m, a1 = ? 2 v12 - u12 = 2a1s1

...(i)

u/2

u A

3 cm

speed = 0 B

C

2

æ uö \ ç ÷ - u 2 = 2 ´ a ´ 0.03 è 2ø

Þ

3 2 u2 - u 2 = 0.06a Þ - u = 0.06a 4 4

8 (c)

9 (d)

10 (a)

11 (c)

12 (b)

13 (b)

14 (a)

15 (c)

-3 u2 4 ´ 0.06 Activity B to C: Assuming the same retardation

Þa =

u2 = u /2 ; v2 = 0 ; s2 = ? ; a2 = v22 - u 22 = 2a2 ´ s2

\ 0-

...(ii)

æ -3 u 2 ö u2 = 2ç ÷ ´ s2 4 è 4 ´ 0.06 ø

Þ s2 =

1 m = 1 cm 100

-3 u2 4 ´ 0.06

EBD_7764 P-8

Physics

Alternatively, dividing (i) and (ii),

v = 0, s = s, a = a

v12 v22

5ö 2 æ Þ 0 - ç100 ´ ÷ = 2as è 18 ø

- u12 - u22

=

2a ´ s1 2a ´ s2

2

2

2

Þ

2.

æ uö 2 çè ÷ø - u 2 2

=

0.03 Þ s2 = 1 cm. s2

æ uö 0-ç ÷ è 2ø (d) For car 1 u1 = u, v1 = 0, a1 = – a, s1 = s1

\ v12 - u12 = 2a1s1 Þ – u2 = – 2as1 ...(i) Þ u2 = 2as1 For car 2 u2 = 4u, v1 = 0, a2 = – a, s2 = s2

5.

\ v22 - u22 = 2a2 s2 Þ – (4u)2 = 2(–a) s2 Þ 16 u2 = 2as2 ...(ii) Dividing (i) and (ii),

5ö æ Þ - ç100 ´ ÷ = 2as … (ii) è 18 ø Dividing (i) and (ii) we get 100 ´ 100 2 ´ a ´ s = Þ s = 24m 50 ´ 50 2´a´6 1 2 (a) We have s = ut + gt , 2 1 2 (Q u = 0) or h = gT 2 now for T/3 second, vertical distance moved is given by 2

h' =

s 1 2as1 = 1 = Þ 2 16 s2 2as2 16u

4.

(b) Ball A is thrown upwards from the building. During u its downward journey A u when it comes back to the point of throw, its speed is equal to the h speed of throw. So, for the journey of both the B balls from point A to B . We can apply v2 – u2 = 2gh. As u, g, h are same for both the balls, vA = vB 5 (c) Case-1 : u = 50 ´ m / s, 18 v = 0,s = 6m, a = a v 2 - u 2 = 2as 2

5ö æ Þ 02 - ç 50 ´ ÷ = 2 ´ a ´ 6 è 18 ø 2

5ö æ Þ - ç 50 ´ ÷ = 2 ´ a ´ 6 ....(i) è 18 ø 5 m/sec Case-2 : u = 100 km/hr = 100 ´ 18

1 æTö 1 gT 2 h g ç ÷ Þ h' = ´ = 2 è 3ø 2 9 9

\ position of ball from ground = h -

u2

3.

\ v 2 - u 2 = 2as

=

6.

7.

(d)

h 9

8h 9

5 50 m/s = m/s 18 3 5 100 d = 20m, u' = 120 ´ = m/s 18 3 2 Let declaration be a then (0) – u2 = –2ad or u2 = 2ad … (1) and (0)2 – u'2 = –2ad'

Speed, u = 60 ´

or u '2 = 2 ad ' …(2) (2) divided by (1) gives, d' 4 = Þ d ' = 4 ´ 20 = 80m d 1 2 (d) Distance from A to B = S = ft1 2 Distance from B to C = ( ft1 ) t Distance from C to D = = ft12 = 2 S A f B t1

t 15 S

( ft1 )2 u2 = 2a 2( f / 2) C f /2 D 2t 1

P-9

Motion in a Straight Line

Þ S + f t1t + 2 S = 15 S ............. (i) Þ f t1t = 12 S 1 2 f t1 = S 2

............ (ii)

Dividing (i) by (ii), we get t1 =

t 6

2

10. (a)

uur change in velocity D v = = time interval t

x

x

a2 2 t 4 dx (c) We know that, v = Þ dx = v dt dt

Þ 2 x = at Þ x =

D v = v 2 + (- v 1 )

11.

90° - v1

t

dx

ò x = a ò dt ; éê 2 x ùú = a[t ]t0 0 0 ë 1 û0

v2 N

W

dv 3 æ dx = vö = f = - 2av çèQ ÷ø dt dt dx dx =a xÞ = a dt v=a x , dt x

Þ

1 ætö f t2 fç ÷ = 2 è 6ø 72 (c) Average acceleration

Þ S=

8.

1 . v Again differentiating, dx 1 dv 2a + 0 = - 2 dt v dt

2ax + b =

v1

E

x

t

0

0

Integrating, ò dx = ò v dt t

or x = ò (v0 + gt + ft 2 ) dt 0

Suur ur ur uur uur v1 = 5iˆ, v2 = 5 ˆj , D v = (v 2 - v 1 ) =

v12 + v22 + 2v1v2 cos 90

= 5 2 + 52 + 0 [As | v1 | = | v2 | = 5 m/s] = 5 2 m/s

uur 5 2 1 Dv = = m / s2 Avg. acc. = 10 t 2 5 tan q = = -1 -5 which means q is in the second quadrant. (towards north-west)

9.

(d)

t

é gt 2 ft 3 ù = êv0 t + + ú 2 3 úû ëê 0 gt 2 ft 3 + 2 3 g f At t = 1, x = v0 + + . 2 3 12. (b) For the body starting from rest

or, x = v0 t +

x1 = 0 +

1 2 at 2

Þ x1 =

1 2 at 2

x1 – x2

t = ax 2 + bx ; Diff. with respect to time (t)

d d dx dx (t ) = a ( x 2 ) + b = a.2 x + b.v. dt dt dt dt 1 = 2axv + bv = v (2ax + b)(v = velocity)

v/a

t

For the body moving with constant speed x2 = vt

EBD_7764 P-10

Physics

1 2 at - vt 2 at t = 0, x1–x2 = 0 v For t < ; the slope is negative a v For t = ; the slope is zero a v For t > ; the slope is positive a These characteristics are represented by graph (b). (b) For downward motion v = –gt The velocity of the rubber ball increases in downward direction and we get a straight line between v and t with a negative slope. \ x1 - x2 =

13.

1 2 Also applying y - y0 = ut + at 2 1 2 1 2 We get y - h = - gt Þ y = h - gt 2 2 The graph between y and t is a parabola with y = h at t = 0. As time increases y decreases. For upward motion. The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here v = u – gt where u is the velocity just after collision. As t increases, v decreases. We get a straight line between v and t with negative slope. 1 2 Also y = ut - gt 2 All these characteristics are represented by

graph (b). 14. (a)

dv = -2.5 v dt

Þ

dv = – 2.5 dt v

Integrating, 0

ò6.25 v

-½

t

dv = -2.5ò dt 0

0

é v +½ ù t = -2.5 [ t ]0 Þ ê (½) ú ëê ûú 6.25

Þ – 2(6.25)½ = – 2.5t Þ t = 2 sec 15. (c) Speed on reaching ground v=

u

u 2 + 2 gh H

Now, v = u + at Þ

u 2 + 2 gh = -u + gt

Time taken to reach highest point is t=

Þ

u , g

t=

u + u 2 + 2 gH

=

nu (from g

g question) Þ 2gH = n(n –2)u2 16. (a) For a body thrown vertically upwards acceleration remains constant (a = – g) and velocity at anytime t is given by V = u – gt During rise velocity decreases linearly and during fall velocity increases linearly and direction is opposite to each other. Hence graph (a) correctly depicts velocity versus time.

3

Motion in a Plane 1.

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10m/s at an angle of 30º with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground ? [2003]

1 3 ] , cos 30o = 2 2 (a) 5.20m (b) 4.33m (c) 2.60m (d) 8.66m The co-ordinates of a moving particle at any time ‘t’are given by x = at3 and y = bt3. The speed of the particle at time ‘t’ is given by [2003]

5.

[ g = 10m/s2 , sin 30o =

2.

3.

4.

2

(a)

3t a + b

2

(b)

(c)

t 2 a 2 + b2

(d)

3t

2

2

a +b

6.

10 s is :

A projectile can have the same range ‘R’ for two angles of projection. If ‘T1’ and ‘T2’ to be time of flights in the two cases, then the product of the two time of flights is directly proportional to. [2004] 1 (a) R (b) R 1 (c) (d) R 2 R2 Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed ? [2004] (a) The acceleration vector points to the centre of the circle (b) The acceleration vector is tangent to the circle (c) The velocity vector is tangent to the circle (d) The velocity and acceleration vectors are perpendicular to each other.

[2009]

(a) 7 2 units (c) 8.5 units

2

a 2 + b2

A ball is thrown from a point with a speed 'v0' at an elevation angle of q. From the same point and at the same instant, a person starts running ' v0 ' with a constant speed to catch the ball. 2 Will the person be able to catch the ball? If yes, what should be the angle of projection q ?[2004] (a) No (b) Yes, 30° (c) Yes, 60° (d) Yes, 45° A particle has an initial velocity of 3iˆ + 4 ˆj and an acceleration of 0.4iˆ + 0.3 ˆj . Its speed after

7.

8.

(b) 7 units (d) 10 units r A particle is moving with velocity n = k ( yiˆ + xjˆ) , where k is a constant. The general equation for its path is [2010] (a) y = x2 + constant (b) y2 = x + constant (c) xy = constant (d) y2 = x2 + constant A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly.[2010] y B P(x,y) m 20

O

A

x

EBD_7764 P-12

Physics

13m/s2

12 m/s2

(a) (b) (c) 7.2 ms2 (d) 14m/s2 For a particle in uniform circular motion, the r acceleration a at a point P(R,q) on the circle of radius R is (Here q is measured from the x-axis ) [2010]

9.

(a)

-

n2 n2 cos q iˆ + sin q ˆj R R

(b)

-

n2 n2 sin q iˆ + cos q ˆj R R

(c)

-

n2 n2 cos q iˆ sin q ˆj R R 2

13. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ? [2015] (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2) (The figures are schematic and not drawn to scale) (a)

2

n ˆ n ˆ i+ j R R A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is : [2011]

240

(y2 – y1) m

(d) 10.

11.

12.

(a)

p

(c)

p

v4 g2 v2 g2

(b)

p v4 2 g2

(d)

p

8 (b) 240

8

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be [2012]

20 2 m

(b) 10 m

(c)

10 2 m

(d) 20 m

(c)

240

12

t(s)

(y2 – y1 ) m

t® 8

A projectile is given an initial velocity of (iˆ + 2 ˆj ) m/s, where iˆ is along the ground and ˆj is along the vertical. If g = 10 m/s2, the

t(s)

(y2 – y1 ) m

v2 g

(a)

12

(d) 240

12

t(s)

(y2 – y1 ) m

equation of its trajectory is : [2013] 2 (a) y = x – 5x (b) y = 2x – 5x2 2 (c) 4y = 2x – 5x (d) 4y = 2x – 25x2

12

t(s)

Answer Key 1 (d)

2 (b)

3 (a)

4 (b)

5 (c)

6 (a)

7 (a)

8 (d)

9 (c)

10 (a)

11 (d)

12 (b)

13 (b)

P-13

Motion in a Plane

(d) From the figure it is clear that range is required

2.

(a)

8.

(d) s = t3 + 5

u 2 sin 2q (10)2 sin(2 ´ 30°) = =5 3 g 10 u 30° Range R

10m

R=

Tower

(b) x = at3 and y = bt3 dx dy vx = = 3at 2 and v y = = 3bt 2 dt dt

Þ velocity, v =

3.

2

4. 5.

6.

v 2 9t 4 = R R At t = 2s, at = 6 × 2 = 12 m/s2 9 ´ 16 ac = = 7.2 m/s2 20 \ Resultant acceleration =

2

4u sin q cos q =2R g

u 2 sin 2 q ) g Hence it is proportional to R. (b) Only option (b) is false since acceleration vector is always radial (i.e., towards the center) for uniform circular motion. (c) Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball’s velocity, the motion of ball will be only in vertical direction with respect to person for that, vo = vo cos q or q = 60° 2 r r (a) Given u = 3iˆ + 4 ˆj , a = 0.4iˆ + 0.3 ˆj , t = 10 s r r r v = u + at = 3iˆ + 4 ˆj + (0.4iˆ + 0.3 ˆj) ´ 10 (Q R =

= 7iˆ + 7 ˆj r \ | v |= 72 + 72 = 7 2 units

dv = 6t dt

Radial acceleration ac =

2

= 3t a + b (a) The angle for which the ranges are same is complementary. Let one angle be q, then other is 90° – q 2u sin q 2u cos q T1 = , T2 = g g

T1T2 =

ds = 3t 2 dt

Tangential acceleration at =

\ v = v 2x + v2y = 9a 2t 4 + 9b2t 4 2

r v = k ( y iˆ + x ˆj ) x-component of v = ky dx = ky ...(1) Þ dt y-component of v = kx dy = kx ...(2) Þ dt dy x From (1) and (2) , dx = y Þ ydy = xdx Þ y2 = x2 + constant

7.

10m

1.

at2 + ac2

(12) 2 + (7.2) 2 = 144 + 51.84 = 195.84 = 14 m/s2 r (c) Clearly a = ac cos q(-iˆ) +ac sin q(- ˆj ) =

9.

=

-v 2 v2 cos q iˆ - sin q ˆj R R Y P( R, q) R O

q

X

10. (a) Total area around fountain 2 A = pRmax

Where Rmax =

v 2 sin 2q v 2 sin 90° v 2 = = g g g

EBD_7764 P-14

\ A= p 11.

(d) R =

v4 g

2

12.

2

2

2

2

u sin q u sin q ,H= 2g g

Hmax at 2q = 90°

u2 Hmax = 2g u2 = 10 Þ u2 = 10 g ´ 2 2g R=

u 2 sin 2q u2 Þ Rmax = g g

Rmax =

10 ´ g ´ 2 = 20 metre g

13.

r (b) From equation, v = iˆ + 2 ˆj Þ x=t 1 y = 2t - (10t 2 ) 2 From (i) and (ii), y = 2x – 5x2 (b) y1 = 10t – 5t2 ; y2 = 40t – 5t2 for y1 = – 240m, t = 8s \ y2 – y1 = 30t for t < 8s. for t > 8s, 1 y2 – y1 = 240 – 40t – gt2 2

Physics

… (i) … (ii)

P-15

Laws of Motion

4

Laws of Motion 1.

2.

3.

4.

5.

6.

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively [2002] (a) g, g (b) g – a, g – a (c) g – a, g (d) a, g When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed then the acceleration of the particle is [2002] (a) F1/m (b) F2F3 /mF1 (c) (F2 - F3)/m (d) F2 /m. The minimum velocity (in ms-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is (a) 60 (b) 30 [2002] (c) 15 (d) 25 A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling) [2002] (a) solid sphere (b) hollow sphere (c) ring (d) all same. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are [2002] (a) 12 N, 6 N (b) 13 N, 5 N (c) 10 N, 8 N (d) 16N, 2N. A light string passing over a smooth light pulley connects two blocks of masses m1 and m 2 (vertically). If the acceleration of the system is g/8, then the ratio of the masses is [2002] (a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3.

7.

Three identical blocks of masses m = 2 kg are drawn by a force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C? [2002] C

8.

B

A

F

(a) 9.2 (b) 3.4 (c) 4 (d) 9.8 One end of a massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms-2) can a man of 60 kg climb on the rope? [2002] P C

9.

(a) 16 (b) 6 (c) 4 (d) 8 A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration

of 5 m / s 2 , the reading of the spring balance will be [2003] (a) 24 N (b) 74 N (c) 15 N (d) 49 N 10. Three forces start acting simultaneously on a r particle moving with velocity, v . These forces are represented in magnitude and direction by the three sides of a triangle ABC. The particle

EBD_7764 P-16

Physics

will now move with velocity C r (a) less than v r (b) greater than v r (c) v in the direction of

11.

[2003]

the largest force BC A B (d) vr , remaining unchanged A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is [2003] (a) 20 N (b) 50 N

10N

(c) 100 N 12.

13.

14.

15.

(d) 2 N A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is [2003] (a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06 A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is[2003] Pm Pm (a) (b) M +m M -m PM (c) P (d) M +m A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is [2003] (a) both the scales read M kg each (b) the scale of the lower one reads M kg and of the upper one zero (c) the reading of the two scales can be anything but the sum of the reading will be M kg (d) both the scales read M/2 kg each A rocket with a lift-off mass 3.5 ´ 10 4 kg is blasted upwards with an initial acceleration of 10m/s2. Then the initial thrust of the blast is (a) 3.5 ´ 10 5 N (b) 7.0 ´ 10 5 N [2003] 5

16.

5

(c) 14.0 ´ 10 N (d) 1.75 ´ 10 N Two masses m1 = 5g and m2 = 4.8 kg tied to a string are hanging over a light frictionless

pulley. What is the acceleration of the masses when left free to move ? ( g = 9.8m / s2 ) [2004] (a)

5 m / s2

(b)

9.8 m / s 2

(c)

0.2 m / s 2

(d)

4.8 m / s 2

17. A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m / s 2 ) [2004] (a) 1.6 (b) 4.0 (c) 2.0 (d) 2.5 18. A smooth block is released at rest on a 45° incline and then slides a distance ‘d’. The time taken to slide is ‘n’ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is [2005] (a)

1 mk = 1 – 2 n

(c)

ms = 1 -

1

(b)

mk = 1-

(d)

ms = 1-

1 n2 1

n n2 19. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ? [2005] (a) 182 m (b) 91 m (c) 111 m (d) 293 m 20. An annular ring with inner and outer radii R1 2

and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the F1 inner and outer parts of the ring , F is[2005] 2

(a)

æ R1 ö çè R ÷ø 2

(c)

R1 R2

2

(b)

R2 R1

(d) 1

Laws of Motion 21. The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by [2005] (a) 2 cos f (b) 2 sin f

22.

(c) tan f (d) 2 tan f A particle of mass 0.3 kg subject to a force F = – kx with k = 15 N/m . What will be its initial acceleration if it is released from a point 20 cm away from the origin ? [2005] (a) 15 m / s 2

23.

(b) 3 m / s 2

P-17 horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m. [2007] mF MF (a) (b) M (m + M ) mF (c) ( M + m) F (d) (m + M ) m 28. Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? [2010]

A

(c) 10 m / s 2 (d) 5 m / s 2 A block is kept on a frictionless inclined surface with angle of inclination ‘ a ’ . The incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to [2005]

B

30°

60°

a

24.

25.

26.

27.

a

(a) g cosec a (b) g / tan a (c) g tan a (d) g Consider a car moving on a straight road with a speed of 100 m/s . The distance at which car can be stopped is [ m k = 0.5 ] [2005] (a) 1000 m (b) 800 m (c) 400 m (d) 100 m A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider g = 10 m/s2). [2006] (a) 4 N (b) 16 N (c) 20 N (d) 22 N A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to [2006] (a) 150 N (b) 3 N (c) 30 N (d) 300 N A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The block are kept on a smooth

(a) 4.9 ms–2 in horizontal direction (b) 9.8 ms–2 in vertical direction (c) Zero (d) 4.9 ms–2 in vertical direction 29. The minimum force required to start pushing a body up rough (frictional coefficient m) inclined plane is F1 while the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an angle q from the horizontal such F that tan q = 2m then the ratio 1 is [2011 RS] F2 (a) 1 (b) 2 (c) 3 (d) 4 30. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of length l A : l B = 2 : 3, then the stiffness of spring ‘A’ is given by [2011 RS] (a)

3k 5

(b)

2k 5

5k 2 31. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves? [2012]

(c) k

(d)

EBD_7764 P-18

Physics

33. A block of mass m is placed on a surface with a

F0 mb

(a)

x3 . If the 6 coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: [2014]

vertical cross section given by y =

v(t) t

(b)

F0 mb v(t)

(a)

t F0 mb

34. t

F0 mb

(d) v(t) t

2g 3

(b)

g 2

(c)

2 m 3

1 1 m m (d) 3 2 Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is: [2015]

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall or release? [2014] (a)

(b)

(c)

(c) v(t)

32.

1 m 6

F

B

A

R m

5g 6

(a) 120 N (c) 100 N

(b) 150 N (d) 80 N

(d) g m

Answer Key 1 (c) 16 (c) 31 (c)

2 (a) 17 (c) 32 (b)

3 (b) 18 (b) 33 (a)

4 (d) 19 (d) 34 (a)

5 (b) 20 (c)

6 (b) 21 (d)

7 (b) 22 (c)

8 (c) 23 (c)

9 (a) 24 (a)

10 (d) 25 (d)

11 (d) 26 (c)

12 (d) 27 (d)

13 (d) 28 (d)

14 (a) 29 (c)

15 (b) 30 (d)

P-19

Laws of Motion

1.

2.

3.

4.

(c) •

For the man standing in the left, the acceleration of the ball r r r abm = ab - am Þ abm = g – a Where 'a' is the acceleration of the mass (because the acceleration of the lift is 'a' ) • For the man standing on the ground, the acceleration of the ball r r r abm = ab - am Þ abm = g – 0 = g (a) When F1 , F2 and F3 are acting on a particle then the particle remains stationary. This means that the resultant of F1 , F2 and F3 is zero. When F1 is removed, F2 and F3 will remain. But the resultant of F2 and F3 should be equal and opposite to F1. i.e. |F2 + F3| = |F1| | F2 + F3 | F1 \ a= Þ a= m m (b) For negotiating a circular curve on a levelled road, the maximum velocity of the car is vmax = mrg Here m = 0.6, r = 150 m, g = 9.8 \ vmax = 0.6´150´9.8 ; 30m / s (d) This is a case of sliding (the plane being frictionless) and therefore the acceleration of all the bodies is same (g sin q).

6.

m1

g 8 m1 -1 1 m2 \ = 8 m1 +1 m2

Here a =

Þ

(b)

Þ

(b) Let the two forces be F1 and F2 and let F2 < F1 . R is the resultant force. Given F1 + F2 = 18 ...(i) 2

R =

F12

F = ( m + m + m) ´ a 10.2 m / s2 6

F

T2 T2

B

T1

T1

A

(c) mg – T = ma \ a=g-

360 T = 10 = 4m / s 2 m 60 T a

T

R mg

F1

10.2 = 3.4N 6

®

...(ii) Only option (b) follows equation (i) and (ii). F1

C

8.

m1 m +1 = 8 1 - 8 m2 m2

m1 9 = m2 7

\ T2 = ma = 2 ´

From the figure F22 + F12 - F22 = R 2 \ F12 - F22 = 144

a

m1g Adding the equations we get (m - m2 ) g a= 1 m1 + m2

q

F2

T

m2g

\a =

5.

T

a m2

7.

g sin q

(b) For mass m1 m1g – T = m1a For mass m2 T–m2g = m2a

mg

EBD_7764 P-20

9.

Physics

(a) For the bag accelerating down mg – T = ma

T

49 (10 – 5) = 24.5 N 10 (d) As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order. Therefore the r r resultant force is zero. Fnet = ma. Therefore, acceleration is also zero i.e., velocity remains unchanged. (d) For the block to remain stationary with the wall f=W

\ T = m( g – a ) =

11.

P m+M Taking the block as a system, we get T = Ma \ a=

MP m+M 14. (a) The Earth pulls the block by a force Mg. The block in turn exerts a force Mg on the spring of spring balance S1 which therefore shows a reading of M kgf. The spring S1 is massless. Therefore, it exerts a force of Mg on the spring of spring balance S2 which shows the reading of M kgf. \ T=

a mg

10.

we get P = (m + M) a

s2

Mkgf

Mkgf

s1

f = mN

M 10N

10N

10N

Mg

15. (b) As shown in the figure F – mg = ma Thrust (F)

W

mN = W 0.2 × 10 = W Þ W = 2 N 12.

(d) u = 6 m/s, v = 0, t = 10s, a = ? v = u + at Þ 0 = 6 + a ´ 10 Þ

a=

a

-6 = -0.6m / s2 10 mg

f = mN N The retardation is due to the frictional force \ f = ma Þ mN = ma Þ mmg = ma

13.

a 0.6 Þm= = = 0.06 g 10 (d) Taking the rope and the block as a system a M m T P

mg

\ F = m ( g + a) = 3.5 × 104 ( 10+10) = 7 × 105 N æ m - m2 ö g 16. (c) Acceleration a = ç 1 è m1 + m2 ÷ø

=

(5 - 4.8) ´ 9.8 m / s2 = 0.2 m/s2 (5 + 4.8)

P-21

Laws of Motion 17. (c)

fs

N

50 m mg

v a = - 2 m / s2

30°

mg sinq = f s ( for body to be at rest)

18.

(b)

Þ m ´ 10 ´ sin 30° = 10 Þ m ´ 5 = 10 Þ m = 2.0 kg g sin q - mg cos q

d q n g si d 45° 45° smooth rough When surface is When surface is smooth rough 1 d = ( g sin q)t12 , 2 1 2 d = ( g sin q - mg cos q) t2 2 t1 =

2d , t2 = g sin q

2d 2d = g sin q - mg cos q g sin q m , as applicable here, is coefficient of kinetic friction as the block moves over the inclined plane.

n

1 1- m k

æ 1 ö ççQ cos 45° = sin 45° = ÷÷ 2ø è n2 =

19.

v2 - u2 32 - 980 » 243 m = 4 2´2 Initially he has fallen 50 m. \ Total height from where he bailed out = 243 + 50 = 293 m

S=

20. (c)

a2 a1

R2 R1

v 2 = wR 2 v1 = wR 1

2d g sin q - mg cos q

According to question, t2 = nt1

n=

3m / s The velocity at ground, v = 3m/s

1 1 or 1- m k = 2 1 - mk n

1 or m k = 1 - 2 n (d) Initial velocity of parachute after bailing out, u=

2gh

u=

2 ´ 9.8 ´ 50 = 14 5

a1 =

v12 w 2 R12 = = w 2 R1 R1 R1

v22 = w 2 R2 R2 Taking particle masses equal F1 ma1 a1 R1 = = = F2 ma2 a2 R2 a2 =

The force experienced by any particle is only along radial direction. Force experienced by the particle, F = mw2R F R \ 1 = 1 F2 R2 21. (d) Acceleration of block while sliding down upper half = g sin f ; retardation of block while sliding down lower half = – ( g sin f - mg cos f) For the block to come to rest at the bottom, acceleration in I half = retardation in II half.

EBD_7764 P-22

Physics

Whand + Wgravity = DK

g sin f = -( g sin f - mg cos f) Þ m = 2 tan f

23.

15 -150 x= x = - 50 x 0.3 3

a = – 50 × 0.2 = 10 m / s 2 (c) From free body diagram, g cos ma N a a a a mg cosa + ma sina mg

mg sin a

For block to remain stationary, mg sin a = ma cos a Þ a = g tan a 24.

(a)

v 2 - u 2 = 2 as or

02 - u 2 = 2( -m k g ) s 1 -1002 = 2 ´ - ´ 10 ´ s 2 s = 1000 m 25. (d) Let the velocity of the ball just when it leaves the hand is u then applying, v2 – u2 = 2as for upward journey Þ -u 2 = 2( -10) ´ 2 Þ u 2 = 40 Again applying v2 – u2 = 2as for the upward journey of the ball, when the ball is in the hands of the thrower, v2 – u2 = 2as

F=

N

N m

a T T

M

mg sin q q

f1

mg

N2

mg cos q

f2

Þ 40 - 0 = 2 (a) 0.2 Þ a = 100 m/s2

\ F = ma = 0.2 ´ 100 = 20 N Þ N - mg = 20 Þ N = 20 + 2 = 22 N

F

mg Mg we get T = ma and F – T = Ma where T is force due to spring Þ F – ma = Ma or,, F = Ma + ma F . \ a= M +m Now, force acting on the block of mass m is æ F ö = mF ma = m ç . è M + m ÷ø m + M 28. (d) mg sin q = ma \ a = g sin q where a is along the inclined plane \ vertical component of acceleration is g sin2 q \ relative vertical acceleration of A with respect to B is g g (sin 2 60 - sin 2 30] = = 4.9 m/s2 2 in vertical direction 29. (c) F1 N1

2

a=–

m(v - u ) 0.15(0 - 20) = = 30 N t 0.1 27. (d) Writing free body-diagrams for m & M, M m K F

26. (c)

F

22.

According to work-energy theorem, W = DK = 0 (Since initial and final speeds are zero) \ Workdone by friction + Work done by gravity = 0 l i.e., -( µ mg cos f ) + mg l sin f = 0 2 µ cos f = sin f or µ = 2 tan f or 2 (c) Mass (m) = 0.3 kg Þ F = m.a = – 15 x

Þ F (0.2) + (0.2)(10)(2.2) = 0 Þ F = 22 N

mg sin q q

mg

mg cos q

P-23

Laws of Motion

\ Þ

30.

31.

(d)

(c)

For the upward motion of the body mg sin q + f1 = F1 or, F1 = mg sin q + mmg cos q For the downward motion of the body, mg sin q – f 2 = F2 or F2 = mg sin q – mmg cos q F1 sin q + m cos q = sin q - m cos q F2 tan q + m 2m + m 3m = = =3 tan q - m 2m - m m 2l æ 3l ö Here, l A = , l B = ç ÷ è 5ø 5 k l = k Al A = k B l B æ 2l ö kl = kA ç ÷ è 5ø 5k kA = 2 Given that F(t) = F0 e -bt dv Þ m = F0 e -bt dt F0 -bt dv e = dt m v

or T = ma Þ mg – ma = ma g Þ a= 2 33. (a) At limiting equilibrium, m = tanq

m q

dy x 2 = (from question) dx 2 Q Coefficient of friction m = 0.5

tanq = m =

x2 2 Þ x=+1

\

t

32.

F0 m

R

x3 1 = m 6 6 f2

f1

ée F0 é - e - bt - e -0 ù ê ú = ë û êë -b úû 0 mb F Þ v = 0 é1 - e -bt ù û mb ë (b) From figure,

v=

0.5 =

Now, y =

F ò dv = m0 ò e -bt dt 0 0 -bt ù t

y

(

)

a

34. (a)

F

A

20N

B f1

N

100N

Assuming both the blocks are stationary N= F f1 = 20 N f2 = 100 + 20 = 120N f

T T m a mg Acceleration a = Ra and mg – T = ma From equation (i) and (ii) æ aö T × R = mR2a = mR2 çè ÷ø R

…(i) …(ii)

120N Considering the two blocks as one system and due to equilibrium f = 120N

EBD_7764 P-24

Physics

5

Work, Energy and Power 1.

2.

3.

4.

5.

6.

Consider the following two statements :[2003] A. Linear momentum of a system of particles is zero B. Kinetic energy of a system of particles is zero. Then (a) A does not imply B and B does not imply A (b) A implies B but B does not imply A (c) A does not imply B but B implies A (d) A implies B and B implies A A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is [2003] (a) 0.2 J (b) 10 J (c) 20 J (d) 0.1 J A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm is [2003] (a) 12.50 N-m (b) 18.75 N-m (c) 25.00 N-m (d) 6.25 N-m A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ‘t’ is proportional to [2003] (a) t 3/4 (b) t 3/2 (c) t 1/4 (d) t 1/2 A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to [2004] (a) x (b) ex (c) x2 (d) loge x A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ? [2004] (a) 12 J (b) 3.6 J

7.

8.

(c) 7.2 J (d) 1200 J r r r r A force F = (5i + 3 j + 2k ) N is applied over a particle which displaces it from its origin to the r r r point r = (2i - j )m. The work done on the particle in joules is [2004] (a) +10 (b) +7 (c) –7 (d) +13 A body of mass ‘m’, accelerates uniformly from rest to ‘v1’ in time ‘t1’. The instantaneous power delivered to the body as a function of time ‘t’ is [2004] (a)

mv1t 2 t1

(b)

(c)

mv1t t1

(d)

mv12t t12 mv12t t1

9.

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that [2004] (a) its kinetic energy is constant (b) its acceleration is constant (c) its velocity is constant (d) it moves in a straight line 10. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is [2005] M

(a)

kL2 2M

(b)

(c)

ML2 k

(d) zero

Mk L

Work, Energy & Power 11. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is [2005] (a) 20 m/s (b) 40 m/s

12.

æ x 4 x2 ö V ( x) = ç - ÷ J. 2ø è 4 The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is [2006]

(c) 10 30 m/s (d) 10 m/s A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function of time is given by [2005] (a)

(c) 13.

P-25 (c) 1.25 J (d) 0.5 J 16. The potential energy of a 1 kg particle free to move along the x-axis is given by

mv 2 T

2

.t 2

mv 2

(b)

1 mv 2 2 .t 2 T2

T

2

(a) (c)

17.

.t

1 mv 2 .t 2 T2

(d)

A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass . After collision the lst mass moves with velocity

18.

v

3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2 nd mass after collision. [2005] m A before collision

(a)

15.

v

v

3 Aafter collision

(b) v

2

v (d) 3 3 A bomb of mass 16kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velolcity of the 12 kg mass is 4 ms–1. The kinetic energy of the other mass is [2006] (a) 144 J (b) 288 J (c) 192 J (d) 96 J A particle of mass 100g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is [2006] (a) –0.5 J (b) –1.25 J

(c) 14.

3v

m

19.

20.

3 2

(b)

2

1

(d) 2 2 A 2 kg block slides on a horizontal floor with a speed of 4m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000 N/m. The spring compresses by [2007] (a) 8.5 cm (b) 5.5 cm (c) 2.5 cm (d) 11.0 cm An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range [2008] (a) 200 J - 500 J (b) 2 × 105 J - 3 × 105 J (c) 20, 000 J - 50,000 J (d) 2,000 J - 5, 000 J A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is [2008] (a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J The potential energy function for the force between two atoms in a diatomic molecule is a b approximately given by U(x) = 12 - 6 , where x x a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = éëU ( x = ¥) - U at equilibrium ùû , D is [2010]

(a)

b2 2a

(b)

b2 12a

(c)

b2 4a

(d)

b2 6a

EBD_7764 P-26

21.

22.

23.

24.

Physics

Statement -1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. [2010] Statement -2 : Principle of conservation of momentum holds true for all kinds of collisions. (a) Statement -1 is true, Statement -2 is true ; Statement -2 is the correct explanation of Statement -1. (b) Statement -1 is true, Statement -2 is true; Statement -2 is not the correct explanation of Statement -1 (c) Statement -1 is false, Statement -2 is true. (d) Statement -1 is true, Statement -2 is false. At time t = 0 a particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to [2011 RS] (a) constant (b) t 1 (c) (d) t t This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - I: Apoint particle of mass m moving with speed u collides with stationary point particle of mass M. If the maximum energy loss æ 1 2ö possible is given as f çè mv ÷ø then f = 2 æ m ö ç ÷. èM + mø Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision. [2013] (a) Statement - I is true, Statment - II is true, Statement - II is the correct explanation of Statement - I. (b) Statement-I is true, Statment - II is true, Statement - II is not the correct explanation of Statement - II. (c) Statement - I is true, Statment - II is false. (d) Statement - I is false, Statment - II is true. When a rubber-band is stretched by a distance x, it exerts restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is: [2014]

(a)

aL2 + bL3

(b)

(

1 aL2 + bL3 2

)

1 æ aL2 bL3 ö + ç ÷ 2 çè 2 3 ÷ø 25. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to : [2015] (a) 56% (b) 62% (c) 44% (d) 50% 26. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 : [2016] (a) 44 ´ 104 J (b) 49 ´ 104 J (c) 45 ´ 104 J (d) 46 ´ 104 J 27. A point particle of mass m, moves long the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The value of the coefficient of friction m and the distance x (= QR), are, respectively close to : P [2016] (c)

aL2 bL3 + 2 3

(d)

h = 2m 30° Q Horizontal surface

R

(a) 0.29 and 3.5 m (b) 0.29 and 6.5 m (c) 0.2 and 6.5 m (d) 0.2 and 3.5 m 28. A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its

Work, Energy & Power

intial speed is v0 = 10 ms–1. If, after 10 s, its energy 1 2 is mv0 , the value of k will be: [2017] 8 (a) 10–4 kg m–1 (b) 10–1 kg m–1 s–1 (c) 10–3 kg m–1 (d) 10–3 kg s–1

1 (c) 16 (a)

1.

2.

2 (d) 17 (b)

3 (b) 18 (d)

4 (b) 19 (c)

5 (c) 20 (d)

6 (b) 21 (a)

P-27 29. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 secand will be [2017] (a) 9 J (b) 18 J (c) 4.5 J (d) 22 J

Answer Key 7 8 9 (b) (b) (a) 22 23 24 (c) (d) (c)

(c) Kinetic energy of a system of particle is zero only when the speed of each particles is zero. And if speed of each particle is zero, the linear momentum of the system of particle has to be zero. Also the linear momentum of the system may be zero even when the particles are moving. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero. \ A does not imply B but B implies A. (d) The elastic potential energy

3.

1 ´ 200 ´ 0.001 = 0.1 J 2

(b) k = 5 ´ 103 N/m

(

1 W = k x22 - x12 2 =

)

1 ´ 5 ´ 103 é(0.1)2 - (0.05)2 ù ë û 2

5000 ´ 0.15 ´ 0.05 = 18.75 Nm 2 (b) We know that F × v = Power \ F ´ v = c where c = constant

dv \m ´v =c dt

mdv ö æ çè\ F = ma = ÷ dt ø

12 (b) 27 (a)

v

t

0

0

13 (d) 28 (a)

2c 1 2 dx = ´t dt m

\

x

ò dx =

\

14 (b) 29 (c)

\

15 (b)

1 2 mv = ct 2

2c 1 2 ´t m

\ v=

0

where v =

dx dt

t

1 2c ´ ò t 2 dt m

0

3

2c 2t 2 3 ´ Þ xµt 2 m 3 (c) Given : retardation µ displacement i.e., a = - x x=

5.

But a = v

=

4.

11 (b) 26 (b)

\ mò vdv = c ò dt

1 = ´ Force ´ extension 2 =

10 (b) 25 (a)

dv dx

vdv = -x Þ dx

\

ò

v1

x

v dv = - ò xdx 0

(v

- v12 = -

Þ

1 1 æ - x2 ö m v22 - v12 = m ç ÷ 2 2 è 2 ø

2 2

)

v2

(

x2 2

)

\ Loss in kinetic energy, \ DK µ x 2

EBD_7764 P-28

6.

Physics

Using conservation of energy,

(b) Mass of over hanging chain 4 m¢ = ´ (0.6)kg 2 Let at the surface PE = 0 C.M.of hanging part = 0.3 m below the table

7.

8.

4 U i = - m ¢gx = - ´ 0.6 ´ 10 ´ 0.30 2 DU = m ' gx = 3.6J = Workdone in putting the entire chain on the table. (b) Workdone in displacing the particle, rr W = F . x = (5iˆ + 3 ˆj + 2kˆ).(2iˆ - ˆj )

= 10 – 3 = 7 joules (b) Let acceleration of body be a \ v1 = 0 + at1 Þ a =

v1 t1

æ1 2 ö m (10 × 100) = m çè v + 10 ´ 20÷ø 2 or

1 2 v = 800 or v = 1600 = 40 m/s 2

Loss in potential energy = gain in kinetic energy 1 m ´ g ´ 80 = mv 2 2 1 2 10 ´ 80 = v 2 v2 = 1600 or v = 40 m/s 12. (b) u = 0; v = u + aT; v = aT Instantaneous power = F × v = m. a. at = m. a2 . t

vt \ v = at Þ v = 1 t1 rr r r Pinst = F .v = (ma ).v

\ Instantaneous power = m

1 1 Mv 2 = k L2 2 2 Þv=

3

13. (d)

(a) Work done by such force is always zero since force is acting in a direction perpendicular to velocity. \ from work-energy theorem = DK = 0 K remains constant.

10. (b)

T2 v

2

æ mv ö æ v t ö æv ö = ç 1÷ ç 1 ÷ = m 1 t çè t ÷ø è t1 ø è t1 ø 1

9.

v2

t = (v2 )y

u1 = v

u2 = 0 m m In x-direction : mv + 0 = m (0) + m(v2)x æ v ö In y-direction : 0 + 0 = m ç ÷ + m(v2 ) y è 3ø is

k .L M

Þ (v2 ) y =

M

v 3

and (v2)x = v 2

æ v ö + v2 \ v2 = ç è 3 ÷ø

Momentum = M × v =M×

k .L = M

kM . L

11. (b)

100 mgH

30

Þ v2 =

4 2v v2 = + v2 = v 3 3 3

In x-direction, mv = mv1 cosq ...(1) where v1 is the velocity of second mass

20 1 mv 2 + mgh 2

In y-direction,

0=

mv 3

- mv1 sin q

P-29

Work, Energy & Power

or m1v1 sin q =

mv

...(2)

3

K.E.max . =

v/ 3 v v

m

v1 cosq

q v1 v1 sinq

Squaring and adding eqns. (1) and (2)

v12 = v2 +

14.

v2

Þ v1 =

2

v

3 3 (b) Let the velocity and mass of 4 kg piece be v1 and m1 and that of 12 kg piece be v2 and m2. 16 kg

Situation 1 4 kg = m1 v1

Initial momentum =0

m2 = 12 kg Final momentum v2 = m2v2 – m1v1

Situation 2

Applying conservation of linear momentum m2v2 = m1v1 Þ v1 =

12 ´ 4 = 12 ms -1 4

1 1 \ K .E.1 = m1v12 = ´ 4 ´ 144 = 288 J 2 2

1 1 15. (b) K.E = mv 2 = ´ 0.1 ´ 25 = 1.25 J 2 2

æ1 ö W = - mgh = - ç mv 2 ÷ = -1.25 J è2 ø 1 é ù 2 êëQ mgh = 2 mv by energy conservation úû 16. (a) Velocity is maximum when K.E. is maximum For minimum. P.E., dV = 0 Þ x 3 - x = 0 Þ x = ±1 dx 1 1 1 - =- J 4 2 4 K.E.(max.) + P.E.(min.) = 2 (Given) Þ Min. P.E. =

\ K.E.(max.) = 2 +

1 9 = 4 4

1 2 mvmax . 2

1 9 3 2 ´ 1 ´ vmax . = Þ vmax. = 2 4 2 17. (b) Let the block compress the spring by x before stopping. kinetic energy of the block = (P.E of compressed spring) + work done against friction. Þ

1 1 ´ 2 ´ (4)2 = ´ 10,000 ´ x 2 + 15 ´ x 2 2 10,000 x2 + 30x – 32 = 0 Þ 5000 x 2 + 15 x - 16 = 0

-15 ± (15)2 - 4 ´ (5000)(-16) 2 ´ 5000 = 0.055m = 5.5cm. 18. (d) The average speed of the athelete \ x=

v=

100 = 10 m/s 10

1 2 mv 2 If mass is 40 kg then, \ K.E. =

1 ´ 40 ´ (10) 2 = 2000 J 2 If mass is 100 kg then, K.E. =

1 ´ 100 ´ (10) 2 = 5000 J 2 19. (c) Initial kinetic energy of the system K.E. =

1 1 K.Ei = mu 2 + M (0)2 2 2 1 ´ 0.5 ´ 2 ´ 2 + 0 = 1J 2 For collision, applying conservation of linear momentum m × u = (m + M) × v 2 \ 0.5 ´ 2 = (0.5 + 1) ´ v Þ v = m / s 3 Final kinetic energy of the system is 1 K.E f = (m + M )v 2 2 =

EBD_7764 P-30

Physics

Statement II is a case of perfectly inelastic collision. By comparing the equation given in statement I with above equation, we get

1 2 2 1 (0.5 + 1) ´ ´ = J 2 3 3 3 \ Energy loss during collision æ 1ö = ç1 - ÷ J = 0.67J è 3ø =

20.

dU ( x ) =0 dx

(d) At equilibrium :

-12a

Þ

x11

=

1

-6b

æ 2a ö 6 Þ x=ç ÷ è b ø

x5

\ U at equilibrium =

a æ 2a ö ç ÷ è b ø

2

-

b b2 = and 4a æ 2a ö ç ÷ è b ø

æ M ö æ m ö f =ç è m + M ÷ø instead of çè M + m ÷ø Hence statement I is wrong and statement II is correct. 24. (c) Work done in stretching the rubber-band by a distance dx is dW = F dx = (ax + bx2)dx Integrating both sides,

U ( x= ¥ ) = 0

21.

22.

æ b2 ö b2 \ D = 0 -ç- ÷ = è 4a ø 4a (a) In completely inelastic collision, all energy is not lost (so, statement -1 is true) and the principle of conservation of momentum holds good for all kinds of collisions (so, statement -2 is true) . Statement -2 explains statement -1 correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body. (c) K.E. µ t or K. E. = ct Þ

1 2 mv = ct 2

Þ

p2 = ct (Q p = mv) 2m

Þ p = 2ctm Þ F= Þ F µ

23.

2 cm ´

0

0

aL2 bL3 + 2 3

Y pf = 3 m V m

pi

2v

25. (a)

45°

v

X

2m

Initial momentum of the system pi =

[m(2V)2 ´ m(2V)2 ]

= 2m ´ 2V Final momentum of the system = 3mV By the law of conservation of momentum 2 2mv = 3mV Þ

2 2v = Vcombined 3

1 1 1 2 DE = m1V12 + m2V22 - (m1 + m2 )Vcombined 2 2 2

2 t

1 t

P2 P2 2m 2(m + M)

é P2 1 2 ù = mv ú êQ K.E. = 2m 2 ëê ûú =

L

Loss in energy

1

(d) Maximum energy loss =

2

L

W = ò axdx + ò bx 2 dx =

P é M ù 1 ì M ü = mv 2 í ý ê ú 2m ë (m + M) û 2 îm + M þ

4 5 DE = 3mv2 - mv2 = mv2 = 55.55% 3 3 Percentage loss in energy during the collision ; 56%

26. (b) n =

W mgh ´ 1000 10 ´ 9.8 ´ 1 ´ 1000 = = input input input

Input =

98000 = 49 × 104J 0.2

P-31

Work, Energy & Power

Fat used =

49 ´ 10 4 7

5

= 12.89 × 10–3kg.

3.8 ´ 10 27. (a) Loss in P.E. = Work done against friction from p ® Q + work done against friction from Q ® R mgh = m(mgcosq) PQ + mmg (QR) h = m cos q × PQ + m(QR) 2 = m × × + mx 2 = m + mx ..... (i) [sin 30° = ] Also work done P ® Q = work done Q ® R \ m = mx

28.

(a) Let Vf is the final speed of the body. From questions, 1 1 mV f2 = mV02 2 8

æ dV F = mç è dt

Þ

ö 2 ÷ = -kV \ ø

Vf =

V0 = 5 m/ s 2

dV (10–2) = –kV2 dt

ò

dV

10 V

2

10

= -100 K ò dt 0

1 1 - = 100K (10) 5 10

29. (c) Using, F = ma = m 6t = 1.

dV dt

or, K = 10–4 kgm–1 dV dt

[Q

m = 1 kg given]

1

v

ét 2 ù –1 dV = 6 t dt 6 V = ê ú = 3 ms ò ò 2 ë û0 0 [Q t = 1 sec given] From work-energy theorem,

W = DKE =

(

)

1 1 m V 2 - u 2 = ´ 1 ´ 9 = 4.5 J 2 2

EBD_7764 P-32

Physics

6

System of Particles and Rotational Motion 1.

Initial angular velocity of a circular disc of mass M is w1. Then two small spheres of mass m are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc? [2002] æ M + mö æ M + mö (a) çè (b) çè ÷w ÷w M ø 1 m ø 1 (c)

2.

3.

4.

æ M ö çè ÷w M + 4m ø 1

(d)

æ M ö çè ÷ w1. M + 2mø

Two identical particles move towards each other with velocity 2v and v respectively. The velocity of centre of mass is [2002] (a) v (b) v/3 (c) v/2 (d) zero Moment of inertia of a circular wire of mass M and radius R about its diameter is [2002] (a) MR2/2 (b) MR2 (c) 2MR2 (d) MR2/4. A particle of mass m moves along line PC with velocity v as shown. What is the angular momentum of the particle about P? [2002] C L P l

5.

6.

7.

8.

9.

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is [2003] (a) L (b) 2 L 4 L (c) 4 L (d) 2 r Let F be the force acting on a particle having r r position vector r , and T be the torque of this force about the origin. [2003] r rThen r r (a) r .T = 0 and F .T ¹ 0 r r r r (b) r .T ¹ 0 and F .T = 0 r r r r (c) r .Tr ¹ 0 and Fr .Tr ¹ 0 r (d) r .T = 0 and F .T = 0 A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected ? [2004] (a) Angular velocity (b) Angular momentum (c) Moment of inertia (d) Rotational kinetic energy One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively IA and IB such that [2004] (a)

O

(a) mvL (b) mvl (c) mvr (d) zero. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius t 4R is made from an iron plate of thickness . 4 Then the relation between the moment of inertia I X and IY is [2003] (a)

ΙY = 32 Ι X

(b)

ΙY = 16 Ι X

(c)

ΙY = Ι X

(d)

ΙY = 64 Ι X

I A < IB

(b)

I A > IB

IA dA = I B dB where dA and dB are their densities. 10. A body A of mass M while falling vertically downwards under gravity breaks into two parts; 1 2 a body B of mass M and a body C of mass 3 3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards [2005]

(c)

IA = IB

(d)

P-33

System of Particles and Rotational Motion

11.

12.

(a) does not shift (b) depends on height of breaking (c) body B (d) body C The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is [2005] 1 2 Mr (a) (b) Mr 2 4 5 1 Mr 2 (c) (d) Mr 2 2 A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force uur ‘ F ’ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C. [2005] l B

A P

2l

F

15. A force of – Fkˆ acts on O, the origin of the coordinate system. The torque about the point (1, –1) is [2006] Z

O

Y

X (a) F (iˆ - ˆj ) (b) - F (iˆ + ˆj ) (c) F (iˆ + ˆj ) (d) - F (iˆ - ˆj ) 16. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity w. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity w' = [2006] (a)

w (m + 2 M ) m

(b)

w(m - 2 M ) (m + 2 M )

wm wm (d) (m + M ) (m + 2M ) 17. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is a / R form the centre of the bigger disc. The value of a is [2007] (a) 1/4 (b) 1/3 (c) 1/2 (d) 1/6 18. A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. Then its acceleration is [2007] (c)

13.

14.

C

2 l 3 4 l (c) l (d) 3 Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the centre of mass at the same position? [2006] m2 m1 (a) m d (b) m + m d 1 1 2 m1 (c) m d (d) d 2 Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is [2006]

(a)

3 l 2

(b)

(a)

2ml 2

(b)

(c)

3ml 2

(d)

3ml2 ml 2

(a) (c)

g sin q 2

1 - MR / I g sin q 2

(b) (d)

g sin q 1 + I / MR 2 g sin q

1 + MR / I 1 - I / MR 2 19. Angular momentum of the particle rotating with a central force is constant due to [2007] (a) constant torque (b) constant force (c) constant linear momentum (d) zero torque

EBD_7764 P-34

20.

Physics

For the given uniform square lamina ABCD, whose centre is O, [2007] D

F

C

(a)

O A

B

E

I AC = 2 I EF (b) 2 I AC = I EF (c) I AD = 3I EF (d) I AC = I EF A thin rod of length ‘L’ is lying along the x-axis with its ends at x = 0 and x = L. Its linear density n æ xö (mass/length) varies with x as k ç ÷ , where n è Lø can be zero or any positive number. If the position xCM of the centre of mass of the rod is plotted against ‘n’, which of the following graphs best approximates the dependence of xCM on n? [2008] (a)

21.

xCM

(a)

(b)

(c)

(d) 22.

L L 2 O xCM L 2 O xCM L L 2 O xCM L L 2 O

23. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is w. Its centre of mass rises to a maximum height of [2009]

n

n

n

n

Consider a uniform square plate of side ‘a’ and mass ‘M’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is [2008] 5 1 Ma 2 Ma 2 (a) (b) 6 12 7 2 Ma 2 Ma 2 (c) (d) 12 3

1 lw 6 g

(b)

1 l 2w 2 2 g

1 l 2w2 1 l 2w 2 (d) 3 g 6 g 24. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is: [2011] (c)

(a) g (c)

g 3

(b) (d)

2 g 3 3 g 2

25. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc. [2011] (a) continuously decreases (b) continuously increases (c) first increases and then decreases (d) remains unchanged 26. A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2 ) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2 the number of rotations made by the pulley before its direction of motion is reversed, is: [2011] (a) more than 3 but less than 6 (b) more than 6 but less than 9 (c) more than 9 (d) less than 3 27. A loop of radius r and mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the centre of the loop is zero.What will be the velocity of the centre of the loop when it ceases to slip ? [2013] (a) (c)

rw0 4 rw0 2

(b)

rw0 3

(d) rw0

System of Particles and Rotational Motion P-35 28. A bob of mass m attached to an inextensible moving from B to C. string of length l is suspended from a vertical ur mv Rk$ when the particle is moving support. The bob rotates in a horizontal circle with (b) L = 2 an angular speedw rad/s about the vertical. About from D to A. the point of suspension: [2014] ur mv $ (a) angular momentum is conserved. Rk when the particle is moving (c) L = – (b) angular momentum changes in magnitude 2 from A to B. but not in direction. (c) angular momentum changes in direction but ur éR ù - a ú k$ when the particle is not in magnitude. (d) L = mv êê úû ë 2 (d) angular momentum changes both in moving from C to D. direction and magnitude. 32. A roller is made by joining together two cones 29. Distance of the centre of mass of a solid uniform at their vertices O. It is kept on two rails AB and cone from its vertex is z0. If the radius of its base is CD, which are placed asymmetrically (see figure), with its axis perpendicular to CD and its R and its height is h then z0 is equal to : [2015] 2 centre O at the centre of line joining AB and Cd 5h 3h (see figure). It is given a light push so that it (a) (b) 8 8R starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller 3h h2 (c) (d) will tend to : [2016] 4 4R B 30. From a solid sphere of mass M and radius R a D cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is: 4MR 2 4MR 2 (a) (b) [2015] O 9 3p 3 3p

MR 2

(c) 31.

MR 2

(d)

32 2p 16 2p A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure : [2016]

y D

a

C

V

O

a V V a V A B a V 45° R

(a) 1 a

Which of the following statements is false for ur the angular momentum L about the origin? (a)

C A (a) go straight. (b) turn left and right alternately. (c) turn left. (d) turn right. 33. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l /R such that the moment of inertia is minimum ? [2017]

ur éR ù L = mv ê + a ú k$ when the particle is ê úû ë 2

(b) 3 2

3 2

3 2 34. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held (c)

(d)

EBD_7764 P-36

Physics

vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle q with the vertical is [2017]

1 (c) 16 (d) 31 (a)

1.

2 (c) 17 (b) 32 (c)

3 (a) 18 (b) 33 (c)

5 (d) 20 (d)

6 (a) 21 (a)

2

1 MR 2 2 MR 2 + 2mR 2

´ w1 =

3.

2g cos q 3l

(c)

3g sin q 2l

(d)

2g sin q 2l

10 (a) 25 (c)

11 (c) 26 (a)

12 (d) 27 (c)

13 (c) 28 (c)

14 (c) 29 (d)

15 (c) 30 (a)

(a) M. I of a circular wire about an axis nn' passing through the centre of the circle and perpendicular to the plane of the circle = MR2 Z Y

X

n'

As shown in the figure, X-axis and Y-axis lie in the plane of the ring . Then by perpendicular axis theorem IX + IY = IZ Þ 2 IX = MR2 [Q IX = IY (by symmetry) and IZ = MR2]

M w1 M + 4m

m v + m2 v2 vc = 1 1 m1 + m2 m v

2v m(2v ) + m(-v ) v = = m+m 2

(b)

n

(c) The velocity of centre of mass of two particle system is given by

m

3g cos q 2l

Answer Key 7 8 9 (d) (b) (a) 22 23 24 (d) (c) (b)

(c) When two small spheres of mass m are attached gently, the external torque, about the axis of rotation, is zero and therefore the angular momentum about the axis of rotation is constant. I \ I1w1 = I 2 w 2 Þ w 2 = 1 w1 I2 1 2 Here I1 = MR 2 1 2 2 and I 2 = MR + 2mR 2 \ w2 = 1

2.

4 (d) 19 (d) 34 (c)

(a)

1 MR 2 2 (d) Angular momentum (L) = (linear momentum) × (perpendicular distance of the line of action of momentum from the axis of rotation) = mv × r [Here r = 0 because the line of = mv × 0 action of momentum passes =0 through the axis of rotation]

\ IX =

4.

P-37

System of Particles and Rotational Motion

5.

(d) We know that density (d ) =

mass( M ) volume(V )

\ M = d ´ V = d ´ ( pR 2 ´ t ) . The moment of inertia of a disc is given by 1 I = MR 2 2 1 pd \ I = ( d ´ pR 2 ´ t ) R 2 = t ´ R4 2 2

I t R4 \ X = X X IY tY RY4

6.

7.

=

t ´ R4

Þ 2I =

1 = 64

t ´ (4 R )4 4 1 2 L (a) K .E . Iw , but L = Iw Þ I = 2 w 1L 1 \ K .E. = ´ w 2 = Lw 2w 2 K .E L´w K .E L´w \ = Þ = K .E ' L ' ´ w ' K .E L '´ 2w 2 L \ L' = 4 r r r (d) We know that t = r ´ F F

t

\ I A < IB 10. (a) Does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes into play while breaking. 11. (c) The disc may be assumed as combination of two semi circular parts. Let I be the moment of inertia of the uniform semicircular disc

r

12. (d)

2Mr 2 Mr 2 ÞI= 2 2

l

A

B

(0, 2 l) F

(0, l)

P

2l y

(0, 0) C

uur To have linear motion, the force F has to be applied at centre of mass. i.e. the point ‘P’has to be at the centre of mass y=

m1 y1 + m2 y2 m ´ 2 l + 2m ´ l 4l = = m1 + m2 3m 3

13. (c) Initially,

8. 9.

r r The angle between t and r is 90° and the r r angle between t and F is also 90°. We also know that the dot product of two vectors which have an angle of 90° between them is zero. Therefore (d) is the correct option. (b) Angular momentum will remain the same since external torque is zero. (a) The moment of inertia of solid sphere A 2 2 about its diameter I A = MR . 5 The moment of inertia of a hollow sphere B 2 2 about its diameter I B = MR . 3

m1 ( - x1 ) + m2 x2 Þ m1 x1 = m2 x2 ...(1) m1 + m2 Finally, The centre of mass is at the origin x1– d x2– d ¢ d d¢ m1 m2 O

0=

\0 =

m1 (d - x1 ) + m2 ( x2 - d ') m1 + m2

Þ 0 = m1d - m1 x1 + m2 x2 - m2 d ' Þd'=

m1 d m2

[From (1).]

EBD_7764 P-38

14.

Physics

(c) n

l

D

A

2 l/

C

O B

n' Inn' = M.I due to the point mass at B + M.I due to the point mass at D + M.I due to the point mass at C. 2

æ l ö I nn ' = 2 ´ m ç + m( 2l) 2 è 2 ÷ø

15.

= ml 2 + 2ml 2 = 3ml 2 ur ur uur (c) Torque t = r ´ F = (iˆ - ˆj ) ´ ( - Fkˆ) = F [- iˆ ´ kˆ + ˆj ´ kˆ]

= F ( ˆj + iˆ) = F ( iˆ + ˆj) 16.

éSince kˆ ´ iˆ = ˆj and ˆj ´ kˆ = iˆ ù ë û (d) Applyin g conservation of an gular momentum I'w' = Iw

(mR 2 + 2MR 2 )w ' = mR 2 w

17.

é m ù Þ w' = wê ë m + 2 M úû (b) Let the mass per unit area be s. Then the mass of the complete disc

( 4psR ) ´ 0 + ( -psR ) R = 2

xc.m

2

4psR 2 - psR 2

\ xc.m =

-psR 2 ´ R

3psR2 R 1 \ xc.m = Þa= 3 3 18. (b) This is a standard formula and should be memorized. g sin q a= I 1+ MR 2 uur uur d Lc 19. (d) We know that t c = dt uur where tc torque about the center of mass uur of the body and Lc = Angular momentum about the center of mass of the body. Central forces act along the center of mass. Therefore torque about center of mass is zero. uur uur When t c = 0 then Lc = constt. 20. (d) By the theorem of perpendicular axes, Iz = Ix + Iy or, Iz = 2 Iy (Q Ix = Iy by symmetry of the figure) Z

= s [p(2 R) 2 ] = 4psR 2

Y F

D

C

2R O

The mass of the r emoved disc = s (pR 2 ) = psR 2 Let us consider the above situation to be a complete disc of radius 2R on which a disc of radius R of negative mass is superimposed. Let O be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as : 2 R 4ps R 2 O –ps R

X

O

R A

E

B

Iz ...(i) 2 Again, bythe same theorem Iz = IAC + IBD = 2 IAC (\ IAC = IBD by symmetry of the figure)

\ I EF =

Iz ...(ii) 2 From (i) and (ii), we get, IEF = IAC. 21. (a) · When n = 0, x = k where k is a constant. This means that the linear mass density is constant. In this case the centre of mass

\ I AC =

P-39

System of Particles and Rotational Motion

will be at the midelle of the rod ie at L/2. Therefore (c) is ruled out · n is positive and as its value increases, the rate of increase of linear mass density with increase in x increases. This shows that the centre of mass will shift towards that end of the rod where n = L as the value of n increases. Therefore graph (b) is ruled out.

DB 2a a = = 2 2 2 According to parallel axis theorem

Also, DO =

2

Ma 2 Ma 2 æ a ö Imm ' = I nn ' + M ç = + ÷ è 2ø 6 2

n

æ xö · The linear mass density l = k çè ÷ø L x Here £1 L With increase in the value of n, the centre of mass shift towards the end x = L such that first the shifting is at a higher rate with increase in the value of n and then the rate decreases with the value of n. These characteristics are represented by graph (a). L

L

ò xdm xCM =

0 L

L

ò x (l dx) 0

=

=

L

ò dm

ò l dx

0

0

=

é x n+ 2 ù kê nú êë (n + 2) L úû 0 L

é k x n +1 ù ê nú êë (n + 1) L úû 0

=

0

L (n + 1) n+2

L ; n = 1, 2 2L 3L xCM = ; n = 2, xCM = ;.... 3 4 Ma 2 1 (d) Inn' = M (a 2 + a 2 ) = 12 6 For n = 0 , xCM =

22.

n

m

n

æxö ò k çè L ÷ø dx

L

23. (c)

Ma 2 + 3Ma 2 2 = Ma 2 6 3 O

C. M

h

The moment of inertia of the rod about O is 1 2 ml . The maximum angular speed of the 3 rod is when the rod is instantaneously vertical. The energy of the rod in this 1 condition is I w 2 where I is the moment of 2 inertia of the rod about O. When the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where h is the maximum height to which the centre of mass (C.M) rises 1 2 1 æ 1 2ö 2 \ mgh = I w = 2 çè 3 ml ÷ø w 2 l 2 w2 Þ h= 6g 24. (b) For translational motion, mg – T = ma .....(1) For rotational motion, T.R = I a = I m

A

Reference level for P.E.

C. M

n

æ xö ò k çè L÷ø .xdx 0 L

=

a R

....(2)

R

D

T m

O

Solving (1) & (2), B

C n

m

1

mg

EBD_7764 P-40

Physics a=

25.

26.

mg mg 2 mg 2 g = = = 2 I ö 3m 3 æ mR m + çè ÷ m+ 2 R2 ø 2R

(c) As insect moves along a diameter, the effective mass and hence the M.I. first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases. (a) F = 20t – 5t2 \a = Þ

=

dw = 4t - t 2 dt w

t

0

0

Þ w = 2t 2 q

ò

0

(

)

6

t3 (as w = 0 at t = 0, 6s) 3

dq = ò

0

Þ n=

36 r) R \ P1 > P2 hence air moves from smaller bubble to bigger bubble. (c) Water fills the tube entirely in gravityless condition i.e., 20 cm. 2

5.

(c) Terminal velocity, vT =

vT

9h

(10.5 - 1.5) 9 Þ vT = 0.2 ´ 2 0.2 (19.5 - 1.5) 18 \ vT = 0.1 m/s 2 (a) The condition for terminal speed (vt) is Weight = Buoyant force + Viscous force B=Vr2 g Fv 2

6.

2r ( d1 - d 2 ) g

=

8.

liquid. Therefore we can conclude that r1 < r2 Also r3 < r2 otherwise the ball would have sink to the bottom of the jar. Also r3 > r1 otherwise the ball would have floated in liquid 1. From the above discussion we conclude that r1 < r3 < r2. (c) In case of water, the meniscus shape is concave upwards. Also according to

2T cos q rrg The surface tension (T) of soap solution is less than water. Therefore rise of soap solution in the capillary tube is less as compared to water. As in the case of water, the meniscus shape of soap solution is also concave upwards. 9. (b) Oil will float on water so, (2) or (4) is the correct option. But density of ball is more than that of oil,, hence it will sink in oil. 10. (c) W = 2T DV W = 2T4p[(52) – (3)2] × 10–4 = 2 × 0.03 × 4p [25 – 9] × 10–4 J = 0.4p × 10–3 J = 0.4p mJ 11. (c) From Bernoulli's theorem, ascent formula h =

1 1 P0 + rv12rgh = P0 + rv22 + 0 2 2

v2 = v12 + 2 gh = 0.16 + 2 ´ 10 ´ 0.2

= 2.03 m/s From equation of continuity A2v2 = A1v1 p

D22 D2 ´ v2 = p 1 v1 4 4

Þ D2 = D1 W=V r 1g

\ V r1 g = V r2 g + kvt2 Vg ( r1 - r 2 ) k (d) From the figure it is clear that liquid 1 floats on liquid 2. The lighter liquid floats over heavier \ vt =

7.

12.

v1 = 3.55 × 10–3 m v2

(c) Sum of volumes of 2 smaller drops = Volume of the bigger drop 4 4 2. pr 3 = pR 3 Þ R = 21/ 3 r 3 3 2 Surface energy = T .4pR

= T 4p 22 / 3 r 2 = T .28 / 3 pr 2 .

EBD_7764 P-54

13.

Physics

(a) V rg = 6phrv + V rl g

Þ Vg ( r - rl ) = 6phrv

(

)

' Also Vg r- ri = 6ph ' rv '

\ v 'h' =

R q

(r - rl' ) ´ vh (r - r l )

rq

Access pressure in air bubble =

(7.8 - 1.2) 10 ´ 8.5 ´ 10-4 ´ (7.8 - 1) 13.2 \ v ' = 6.25 ´ 10-4 cm/s (d) The surface tension of the liquid film is F given as T = , where F is the force, and 2l l = length of the slider. F = 2lT At equilibrium, F = W 2Tl = mg

2T R

2T 4p R 3 (p r 2 ) = rw g 3T R

=

14.

T×dl

Force due to excess pressure = upthrust

(r - rl' ) vh ´ Þ v' = (r - rl ) h '

2 Þr =

2 R 4 rw g 2 2rw g Þ r=R 3T 3T

17. (a) 8 cm

mg 1.5 ´ 10 -2 1.5 = = 2 2 l 2 ´ 30 ´ 10 60 = 0.025 Nm–1 (c) Pressure at interface A must be same from both the sides to be in equilibrium.

54 cm

P

(54–x) x

T=

15.

d2

R a a Rcosa

Rsin a – Rsin a

Rsina

R

d1

\

A

( R cos a + R sin a )d 2 g = ( R cos a - R sin a )d1 g

16.

Hg

d1 cos a + sin a 1 + tan a Þ d = cos a - sin a = 1 - tan a 2 (a) When the bubble gets detached, Buoyant force = force due to surface tension

18. (c) \

Q

Length of the air column above mercury in the tube is, P + x = P0 Þ P = (76 – x) Þ 8 × A × 76 = (76 – x) × A × (54 – x) \ x = 38 Thus, length of air column = 54 – 38 = 16 cm. As linear dimension increases by a factor of 9 vf = 93 vi Density remains same So, mass µ Volume mf ( Area ) f = 93 = 92 Þ mi ( Area ) i force (mass ) ´ g Stress (s) = = area area 3 s 2 æ m f ö æ Ai ö 9 = =9 ç ÷ = s1 çè mi ÷ø è A f ø 92

10

Thermal Properties of Matter 1.

2.

3.

4.

5.

6.

Heat given to a body which raises its temperature by 1°C is [2002] (a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient Infrared radiation is detected by [2002] (a) spectrometer (b) pyrometer (c) nanometer (d) photometer Which of the following is more close to a black body? [2002] (a) black board paint (b) green leaves (c) black holes (d) red roses If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should [2002] (a) increase (b) remain unchanged (c) decrease (d) first increase then decrease Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is [2002] (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9. The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by [2003] (a) Rayleigh Jeans law (b) Planck’s law of radiation (c) Stefan’s law of radiation (d) Wien’s law

7.

8.

9.

According to Newton’s law of cooling, the rate of cooling of a body is proportional to (Dq)n, where Dq is the difference of the temperature of the body and the surroundings, and n is equal to [2003] (a) two (b) three (c) four (d) one If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be [2004] (a) 32 (b) 16 (c) 4 (d) 64 The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1 ) . The rate of heat transfer through the slab, in a steady state is æ A(T2 - T1 ) K ö çè ÷ø f , with f equal to x x

K

4x

T1

2K

2 3

(b)

1 2

(c) 1

(d)

1 3

(a)

[2004]

EBD_7764 P-56

10.

Physics

The figure shows a system of two concentric spheres of radii r1 and r2 are kept at temperatures T1 and T2, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to [2005] r1 r2

11.

ær ö ln ç 2 ÷ è r1 ø

(b)

(r2 - r1 ) (r1 r2 )

(c)

( r2 - r1 )

(d)

r1 r2 (r2 - r1 )

T4 4pr02 R 2s 2 r

r02 R2 s

T

(b)

4 2

(d)

T4 pr02 R 2s 2 r

2

R s

T

l2

K1

T2

K2

(a)

( K1l1T1 + K 2 l2T2 ) ( K1l1 + K 2 l2 )

(b)

( K 2l2T1 + K1l1T2 ) ( K1l1 + K 2 l2 )

(c)

( K 2l1T1 + K1l2T2 ) ( K 2 l1 + K1l2 )

( K1l2T1 + K 2l1T2 ) ( K1l2 + K 2 l1 ) 14. A long metallic bar is carrying heat from one of its ends to the other end under steady–state. The variation of temperature q along the length x of the bar from its hot end is best described by which of the following figures? [2009]

(d)

Assuming the Sun to be a spherical body of radius R at a temperature of TK, evaluate the total radiant powerd incident of Earth at a distance r from the Sun [2006]

(c)

13.

T2

(a)

(a)

12.

T1

T1 l1

4

q (a)

(b)

x

2

4pr r where r0 is the radius of the Earth and s is Stefan's constant. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box contains æ 7ö one mole of helium at temperature çè ÷ø T0 . The 3 boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf in terms of T0 is [2006] 3 7 (a) T f = T0 (b) T f = T0 7 3 3 5 (c) T f = T0 (d) T f = T0 2 2 One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of length l1 and l2 and thermal conductivities K 1 and K 2 respectively. The temperature at the interface of the two section is [2007]

q

x

q

q

(c)

(d)

x

x

15. 100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K): [2011] (a) 8.4 kJ (b) 84 kJ (c) 2.1 kJ (d) 4.2 kJ 16. The specific heat capacity of a metal at low temperature (T) is given as 3

æ T ö C p (kJK -1kg -1 ) = 32 ç è 400 ÷ø A 100 gram vessel of this metal is to be cooled from 20ºK to 4ºK by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is [2011 RS]

P-57

Thermal Properties of Matter

17.

(a) greater than 0.148 kJ (b) between 0.148 kJ and 0.028 kJ (c) less than 0.028 kJ (d) equal to 0.002 kJ A wooden wheel of radius R is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than 2pR. To fit the ring on the wheel, it is heated so that its temperature rises by DT and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is a, and its Young's modulus is Y, the force that one part of the wheel applies on the other part is : [2012]

R

(c)

0

19.

loge (q – q0) 0

t

loge (q – q0)

0

(b)

(d) t

t

0

(a)

(b)

O

(c)

T q0

t

If a piece of metal is heated to temperature q and then allowed to cool in a room which is at temperature q 0 , the graph between the temperature T of the metal and time t will be closest to [2013]

T q0 O

t

(d)

t

T q0

t O O t 20. Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure. Area of cross - section of each rod = 4 cm2. End of copper rod is maintained at 100ºC where as ends of brass and steel are kept at 0ºC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is: [2014] (a) 1.2 cal/s (b) 2.4 cal/s (c) 4.8 cal/s (d) 6.0 cal/s 21. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with U internal energy per unit volume u = µ T4 and V 1æUö pressure p = ç ÷ . If the shell now undergoes 3è V ø an adiabatic expansion the relation between T and R is : [2015] 1 1 (a) T µ (b) T µ 3 R R

(c) T µ e–R

loge (q – q0)

(a)

loge (q – q 0)

18.

(a) 2pSY aDT (b) SY aDT (c) pSY aDT (d) 2SY aDT A liquid in a beaker has temperature q(t) at time t and q0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge(q – q0) and t is : [2012]

T

(d) T µ e–3R

22. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (a) of the metal of the pendulum shaft are respectively : [2016] (a) 30°C; a = 1.85 × 10–3/°C (b) 55°C; a = 1.85 × 10–2/°C (c) 25°C; a = 1.85 × 10–5/°C (d) 60°C; a = 1.85 × 10–4/°C

EBD_7764 P-58

23.

Physics

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C [2017] (1) 1250°C

(2) 825°C

(3) 800°C

(4) 885° C

24. An external pressure P is applied on a cube at 0oC so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by : [2017] (1) (3)

1 (b) 16 (d)

1. 2. 3. 4.

5.

2 (b) 17 (d)

3 (a) 18 (a)

4 (c) 19 (c)

5 (a) 20 (c)

6 (d) 21 (a)

Answer Key 7 8 9 (d) (d) (d) 22 23 24 (c) (d) (c)

(b) Heat required for raising the temperature of the whole body by 1ºC is called the thermal capacity of the body. (b) Pyrometer is used to detect infra-red radiation. (a) Black board paint is quite approximately equal to black bodies. (c) When water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass, therefore, when water is cooled to ice, its mass decreases. (a) The energy radiated per second is given by E = esT4 A For same material e is same. s is stefan's constant \

F1 T14 A1 T14 4pr12 = = F2 T24 A2 T24 4pr22

8.

(d)

3a PK P 3a K

10 (d)

11 (b)

(2)

3PKa

(4)

P aK

12 (c)

E2 R22 T24 = E1 R12 T14 put R2 = 2R, R1 = R ; T2 = 2T, T1 = T \

E2 (2R )2 (2T )4 = = 64 E1 R 2T 4 (d) The thermal resistance is given by x 4x x 2 x 3x + = + = KA 2KA KA KA KA dQ DT (T2 - T1 ) KA \ = = 3x dt 3x KA 1 ì A(T2 - T1 ) K ü = í 1 ý 3î x þ \f = 3 Þ

9.

10. (d)

T - dT dr

1 = = 4 2 1 (2000) ´ 4 (d) Wein’s law correctly explains the spectrum (d) According to Newton’s law of cooling -

dQ µ (Dq) dt

14 (a)

2 2 4 E = s AT 4 ; A µ R \ E µ R T

(4000)4 ´12

6. 7.

13 (d)

·

T1

r1

T2 r2

r

15 (a)

Thermal Properties of Matter

Consider a shell of thickness (dr) and of radiius (r) and let the temperature of inner and outer surfaces of this shell be T and (T – dT) respectively. dQ = rate of flow of heat through it dt =

P-59 13. (d) Let T be the temperature of the interface. As the two sections are in series, the rate of flow of heat in them will be equal. T2 T1 2 1

K1

KA[(T - dT ) - T ] - KAdT = dr dr

K1 A(T1 - T ) K 2 A(T - T2 ) , = l1 l2 where A is the area of cross-section. or, K1 A(T1 - T )l 2 = K 2 A(T - T2 )l1

\

dT (Q A = 4pr 2 ) dr To measure the radial rate of heat flow, integration technique is used, since the area of the surface through which heat will flow is not constant. 2 = -4pKr

æ dQ ö Then, ç è dt ÷ø

r2

1

ò r2

r1

or

r r dQ µ 1 2 dt (r2 - r1 ) (b) Total power radiated by Sun = sT4 × 4pR2 The intensity of power at earth's surface =

sT 4 ´ 4pR2 4pr 2

Total power received by Earth =

12.

\T =

sT 4 R 2 2

(pr02 )

K1l 2T1 + K 2 l1T2 . K1l 2 + K 2 l1 14. (a) The heat flow rate is given by dQ kA(q1 - q) = dt x

x dQ x dQ Þ q = q1 kA dt kA dt where q1 is the temperature of hot end and q is temperature at a distance x from hot end. The above equation can be graphically represented by option (a). 15. (a) DU = DQ = mcDT = 100 × 10–3 × 4184 (50 – 30) » 8.4 kJ 16. (d) Required work = energy released Þ q -q = 1

ò

Here, Q = mc dT

r (c) Heat lost by He = Heat gained by N2 n1Cv1 DT1 = n2Cv2 DT2 3 é7 R T - Tf 2 ëê 3 0

ù 5 é = R T f - T0 ùû ûú 2 ë

7T0 - 3T f = 5T f - 5T0 Þ 12T0 = 8T f Þ T f = 3 Þ T f = T0 . 2

12 T0 8

K1T1l 2 + K 2T2 l1 K 2 l1 + K1l 2

=

dQ -4pKr1r2 (T2 - T1 ) = dt (r2 - r1 )

\

11.

or, ( K 2 l1 + K1l 2 )T = K1T1l 2 + K2T2 l1

T2

dQ é 1 1 ù ê - ú = -4pK [T2 - T 1 ] dt ë r1 r2 û

K1T1l 2 - K1T l 2 = K 2T l1 - K 2T2 l1

or,

dr = -4 pK ò dT T1

K2

4

æ T3 ö 0.1 ´ 32 ´ ò ç ÷ dT » 0.002 kJ. è 4003 ø 20 Therefore, required work = 0.002 kJ (d) The Young modulus is given as F/S Y= DL / L Here it is given as F Y= ´ 2 pR {L = 2pR} S2 pDR FR or Y = …(i) S.DR =

17.

EBD_7764 P-60

18.

19. 20.

Physics

The coefficient of linear expansion is given DR as a = RDT DR R 1 = a.DT Þ = Þ … (ii) R DR aDT From equation (i) and (ii) F Y= Þ F = Y.S.aDT S.aDT \ The ring is pressing the wheel from both sides, Thus Fnet = 2F = 2YSaDT (a) Newton's law of cooling dq = - k(q - q0 ) dt dq Þ = - kdt (q - q0 ) Integrating Þ log(q - q0 ) = -kt + c Which represents an equation of straight line. Thus the option (a) is correct. (c) According to Newton’s law of cooling, the temperature goes on decreasing with time non-linearly. (c) Rate of heat flow is given by, KA(q1 - q 2 ) l Where, K = coefficient of thermal conductivity l = length of rod and A = area of crosssection of rod

1æUö 21. (a) As, P = ç ÷ 3èVø U = KT 4 V 1 4 So, P = KT 3

But

or

4 3 3 pR T = constant 3 1 Therefore, Tµ R 22.

Time lost/gained per day =

(c)

1 a (40 – q) ´ 86400 2 1 4 = a (q – 20) ´ 86400 2 40 – q On dividing we get, 3 = q – 20 12 =

Brass

0°C 0°C If the junction temperature is T, then QCopper = QBrass + QSteel 0.92 ´ 4(100 - T ) 0.26 ´ 4 ´ (T - 0) + = 46 13

Þ 200 – 2T = 2T + T Þ T = 40°C \

QCopper =

....(ii)

3q – 60 = 40 – q 4q = 100 Þ q = 25°C

23. (d) According to principle of calorimetry, Heat lost = Heat gain 100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45 10 = 1200 + 7650 = 8850

Copper

Steel

.... (i)

10 – 750 = 450 + 7650

100°C

B

1 µ Dq ´ 86400 2

second

Q=

T

uRT 1 = KT 4 [As PV = u RT] V 3

0.12 ´ 4 ´ (T - 0) 12

0.92 ´ 4 ´ 60 = 4.8 cal/s 46

T = 885°C 24. (c) As we know, Bulk modulus DV P DP = Þ V K æ -DV ö çè ÷ V ø V = V0 (1 + gDt) K=

DV = gDt V0

\

P P P = gDt Þ Dt = = K gK 3aK

a

P-61

Thermodynamics

11

Thermodynamics 1.

2.

3.

4.

5.

6.

Which statement is incorrect? [2002] (a) All reversible cycles have same efficiency (b) Reversible cycle has more efficiency than an irreversible one (c) Carnot cycle is a reversible one (d) Carnot cycle has the maximum efficiency in all cycles Even Carnot engine cannot give 100% efficiency because we cannot [2002] (a) prevent radiation (b) find ideal sources (c) reach absolute zero temperature (d) eliminate friction “Heat cannot by itself flow from a body at lower temperature to a body at higher temperature” is a statement or consequence of [2003] (a) second law of thermodynamics (b) conservation of momentum (c) conservation of mass (d) first law of thermodynamics During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP/CV for the gas is [2003] 4 (b) 2 (a) 3 5 3 (c) (d) 3 2 Which of the following parameters does not characterize the thermodynamic state of matter? [2003] (a) Temperature (b) Pressure (c) Work (d) Volume A Carnot engine takes 3 × 106 cal of heat from a reservoir at 627°C, and gives it to a sink at 27°C. The work done by the engine is [2003] (a) 4.2 × 106 J (b) 8.4 × 106 J 6 (c) 16.8 × 10 J (d) zero

7.

Which of the following statements is correct for any thermodynamic system ? [2004] (a) The change in entropy can never be zero (b) Internal energy and entropy are state functions (c) The internal energy changes in all processes (d) The work done in an adiabatic process is always zero. 8. Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volume (V1, V2), and pressure (P1, P2) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be [2004] (a) T1T2(P1V1 + P2V2)/(P1V1T2 + P2V2T1) (b) (T1 + T2)/2 (c) T1 + T2 (d) T1T2(P1V1 + P2V2)/(P1V1T1 + P2V2T2) 9. Which of the following is incorrect regarding the first law of thermodynamics? [2005] (a) It is a restatement of the principle of conservation of energy (b) It is not applicable to any cyclic process (c) It introduces the concept of the entropy (d) It introduces the concept of the internal energy 10. A system goes from A to B via two processes I and II as shown in figure. If DU1 and DU2 are the changes in internal energies in the processes I and II respectively, then [2005] p

II A

B

I v

EBD_7764 P-62

11.

Physics

(a) relation between DU1 and DU2 can not be determined (b) DU1 = DU2 (c) DU2 = DU1 DU 2 < DU1 (d) DU2 > DU1 The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is [2005] T

2T0

15. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be [2008] (a)

T1T2 ( PV 1 1 + P2V2 ) PV T 1 1 2 + P2V2T1

(b)

PV 1 1T1 + P2V2T2 PV 1 1 + P2V2

(c)

PV 1 1T2 + P2V2T1 PV 1 1 + P2V2

(d)

T1T2 ( PV 1 1 + P2V2 ) PV T 1 1 1 + P2V2T2

T0 S0 (a)

1 4

S

2S0 (b)

1 2

2 1 (d) 3 3 The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is (R = 8.3 J mol–1 K–1) [2006] (a) diatomic (b) triatomic (c) a mixture of monoatomic and diatomic (d) monoatomic When a system is taken from state i to state f along the path iaf, it is found that Q =50 cal and W = 20 cal. Along the path ibf Q = 36 cal. W along the path ibf is [2007] a f

(c)

12.

13.

14.

i b (a) 14 cal (b) 6 cal (c) 16 cal (d) 66 cal A Carnot engine, having an efficiency of h = 1/ 10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [2007] (a) 100 J (b) 99 J (c) 90 J (d) 1 J

Directions for questions 16 to 18 : Questions are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram. [2009] 5

2 × 10

A

B

D

C

300K

500K

P (Pa) 1 × 10

5

T

16. Assuming the gas to be ideal the work done on the gas in taking it from A to B is (a) 300 R (b) 400 R (c) 500 R (d) 200 R 17. The work done on the gas in taking it from D to A is (a) + 414 R (b) – 690 R (c) + 690 R (d) – 414 R 18. The net work done on the gas in the cycle ABCDA is (a) 276 R (b) 1076 R (c) 1904 R (d) zero

Thermodynamics 19. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is [2010] (a) 0.5 (b) 0.75 (c) 0.99 (d) 0.25 20. A Carnot engine operating between

P-63 The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is [2013]

1 . When 6

æ 11 ö (d) 4p0v0 ç ÷ p0 v0 è2ø 24. One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement: [2014]

temperatures T1 and T2 has efficiency

T2 is lowered by 62 K its efficiency increases to 1 . Then T1 and T2 are, respectively: 3

21.

(a) 372 K and 310 K (b) 330 K and 268 K (c) 310 K and 248 K (d) 372 K and 310 K Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure. The efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas) [2012] (a) 15.4 % (b) 9.1 % (c) 10.5% (d) 12.5 %

2P0 P0

22.

23.

[2011]

B

2p0 p0

A D v0

(b)

B

800 K

P

A

D

A

B C 2v0 v

æ 13 ö ç ÷ p0 v0 è2ø

(c)

C

V0 2V0 A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be : [2012] (a) efficiency of Carnot engine cannot be made larger than 50% (b) 1200 K (c) 750 K (d) 600 K p

(a) p 0 v 0

400 K

600 k C

V (a) The change in internal energy in whole cyclic process is 250 R. (b) The change in internal energy in the process CA is 700 R. (c) The change in internal energy in the process AB is -350 R. (d) The change in internal energy in the process BC is – 500 R. 25. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways : [2015] (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature 100°C to final temperature 200°C.

EBD_7764 P-64

26.

Physics

Entropy change of the body in the two cases respectively is : (a) ln2, 2ln2 (b) 2ln2, 8ln2 (c) ln2, 4ln2 (d) ln2, ln2 'n' moles of an ideal gas undergoes a process A ® B as shown in the figure. The maximum temperature of the gas during the process will be : [2016] P

P0

1.

2.

2 (c) 17 (a)

27.

3 (a) 18 (a)

4 (d) 19 (b)

2V0

5 (c) 20 (d)

(b)

n=

C - CV C - CP

(c)

n=

CP CV

(d)

n=

C – CP C – CV

Answer Key 7 8 9 (b) (a) (b, c) 22 23 24 (c) (b) (d)

10 (b) 25 (d)

13 (b)

14 (c)

(a) All reversible engines working for the same temperature of source and sink have same efficiencies. If the temperatures are different, the efficiency is different. (c) In Carnot's cycle we assume frictionless piston, absolute insulation and ideal source and sink (reservoirs). The efficiency of

Þ PT From

For h = 1 or 100 %, T2 = 0 K. The temperature of 0 K (absolute zero) can not be obtained . (a) This is a statement of second law of thermodynamics (d)

....(i) P µ T 3 Þ PT -3 = constant But for an adiabatic process, the pressure temperature relationship is given by P1–g Tg = constant

11 (d) 26 (c)

g 1-g

12 (a) 27 (d)

....(ii)

= constt.

(i)

an d

15 (a)

(ii)

g = -3 1- g

3 2 (c) Work is a path function. The remaining three parameters are state function. T2 (273 + 27) (b) h = 1 - T = 1 - (273 + 627) 1 Þ g = -3 + 3g Þ g=

5.

1

4.

9P0 V0 3P0 V0 (d) 2nR 2nR An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) : [2016]

CP - C C - CV

T2 carnot's cycle is given by h = 1 - T

3.

9P0 V0 nR

n=

V

6 (b) 21 (a)

(b)

(a)

B

V0

9P0 V0 2nR

(c)

A

2P0

1 (a) 16 (b)

(a)

6.

300 1 2 = 1- = 900 3 3 W But h = Q W 2 2 2 \ = Þ W = ´ Q = ´ 3 ´ 106 Q 3 3 3 =1-

P-65

Thermodynamics

= 2 ´ 106 cal

7. 8.

= 2 ´ 106 ´ 4.2 J = 8.4 ´ 106 J (b) Internal energy and entropy are state function, they do not depend upon path taken. (a) Here Q = 0 and W = 0. Therefore, from first law of thermodynamics DU = Q + W = 0 \ Internal energy of the system with partition = Internal energy of the system without partition. n1Cv T1 + n2 Cv T2 = (n1 + n2 )Cv T

\

T=

But n1 =

n1T1 + n2T2 n1 + n2 PV PV 1 1 and n2 = 2 2 RT1 RT2

PV PV 1 1 ´ T1 + 2 2 ´ T2 RT1 RT2 \\ T = PV P 1 1 + 2V2 RT1 RT2 T1T2 ( PV 1 1 + P2V2 ) PV 1 1T2 + P2V2T1 (b, c)First law is applicable to a cyclic process. Concept of entropy is introduced by the second law. (b) Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e., DU1 = DU2 =

9. 10.

11.

(d)

1 3 Q1 = T0 S 0 + T0 S0 = T0 S 0 2 2 Q2 = T0(2S0 – S0) = T0S0 and Q3 = 0

h=

T

T0

Q1

Q3

Q2 S0

2S 0

12. (a)

W=

Q2 TS 1 = 1- 0 0 = 3 Q1 3 T S 2 0 0

nRDT 1000 ´ 8.3 ´ 7 Þ -146000 = 1- g 1- g

58.1 58.1 Þ g = 1+ = 1.4 146 146 Hence the gas is diatomic. 13. (b) For path iaf, Q = 50 cal, W = 20 cal a f

or 1 - g = -

i b By first law of thermodynamics, DU = Q - W = 50 – 20 = 30 cal. For path ibf Q = 36 cal W =? or, W = Q – DU (Since, the change in internal energy does not depend on the path, therefore DU = 30 cal) \ W = Q – DU = 36 – 30 = 6 cal. 14. (c) The efficiency (h) of a Carnot engine and the coefficient of performance (b) of a refrigerator are related as

b=

1- h h

Here, h =

1 10

1 10 = 9. \ b= æ 1ö çè ÷ø 10 1-

Also, Coefficient of performance (b) is given Q by b = 2 , where Q 2 is the energy W absorbed from the reservoir.

W Q1 - Q2 = Q1 Q1

2T0

= 1-

S

\ Q2 = 90 J. or, 9 = Q2 10 15. (a) Here Q = 0 and W = 0. Therefore from first law of thermodynamics DU = Q + W = 0 \ Internal energy of the system with partition = Internal energy of the system without partition. n1Cv T1 + n2 Cv T2 = (n1 + n2 )Cv T

n T + n2 T2 \T = 1 1 n1 + n2

EBD_7764 P-66

Physics

But n1 =

PV PV 1 1 and n2 = 2 2 RT1 RT2

For diatomic gas, g =

PV PV 1 1 ´ T1 + 2 2 ´ T2 RT1 RT2 \T= PV P 1 1 + 2V2 RT1 RT2

\ g -1 =

17.

2

T1T2 ( PV 1 1 + P2V2 ) PV 1 1T2 + P2V2T1 (b) A to B is an isobaric process. The work done W = nR (T2 - T1 )

= 2R (500 - 300) = 400 R (a) Work done by the system intheisothermal process P DA is W = 2.303 nRT log10 D PA

T2 Now, efficiency = 1 - T 1

18.

20.

PB PC

= 2.303 ´ 2R ´ 500log

+ (-400R) - 414R

2 ´ 105

1 ´ 105 = 279.2 R

= 693.2 R – 414 R

- 414 R

(b) (V, T1) (32 V, T2) T2 V

Þ T1V

g -1

g -1

= constant

= T2 (32V )g -1 g -1

Þ T1 = (32)

T

5

Again, h2 = 1 -

....(i) T2 - 62 1 = 3 T1

T1 = 372 K and T2 = 21.

5 × 372 = 310 K 6

....(ii)

output work input work Input work = Work done in going A to B + workdone in going B to C and the work done in going C to D. n n Wi = (P0V0 ) + (2P0V0 ) + 2P0V0 2 2 where n = degree of freedom which is 3 for mono-atomic gases like He

(a) The efficiency h =

æ3 3 ö = ç + .2 + 2 ÷ P0 V0 è2 2 ø

T1

We have, TV

T2 T1

Solving (i) and (ii), we get,

– 414R. 2 ´ 105 Therefore, work done on the gas is + 414 R. (a) The net work in the cycle ABCDA is W = W AB + WBC + WCD + WDA

P

h1 = 1 -

2 Þ T =6 1

1 ´ 105

= 400R + 2.303nRT log

19.

T2 3 1 = 1 - = = 0.75. 4 4 4T2 (d) Efficiency of engine

= 1-

= 2.303 ´ 2 R ´ 300 log10

2 5

\ T = (32) 5 .T Þ T1 = 4T2 1 2

=

16.

7 5

.T2

13 æ 3 + 10 ö =ç ÷ P0 V0 = P0 V0 2 è 2 ø and W0 = P0V0 P0V0 2 = 13 13 P0V0 2 Efficiency in % h=

h=

2 200 ´ 100 = ; 15.4% 13 13

Thermodynamics 22. (c) The efficiency of the engine is given as

æ T ö h = ç 1 - 2 ÷ ´ 100 T1 ø è For first case T1 = 500 K; h = 40

25. 26.

-P0 , c = 3P0] V0

T2 ö æ 40 = ç 1 ÷ ´ 100 è 500 ø T2 40 Þ = 1100 500

PV0 + P0V = 3P0V0 But pv = nRT

T2 60 = Þ T2 = 300 K 500 100 For second case :

nRT V0 + P0V = 3P0V0 v \ nRT V0 + P0V2 = 3P0V0 ...(iii)

æ 300 ö 300 60 40 ÷ = ç1= T T 100 è 100 2 2 ø 100 ´ 300 Þ T2 = Þ T2 = 750 K 40 (b) Heat is extracted from the source in path DA and AB is

æ 13ö 3 5 Þ P0V0 + 2 P0V0 = ç ÷ P0V0 è 2ø 2 2 (d) In cyclic process, change in total internal energy is zero. DUcyclic = 0 5R DT DUBC = nCv DT = 1 ´ 2 Where, Cv = molar specific heat at constant volume. For BC, DT = –200 K \ DUBC = –500R (d) The entropy change of the body in the two cases is same as entropy is a state function. (c) The equation for the line is

P

dT =0 dv Differentiating e.q. (iii) by 'v' we get

For temperature to be maximum

nRV0

dT + P0(2v) = 3P0V0 dv

\ nRV0

2Po Po

V=

3V0 3P \ p= 0 2 2

q Vo Vo

2Vo

V

[From (i)]

9Po Vo [From (iii)] 4nR 27. (d) For a polytropic process

\ Tmax =

C = Cv +

\ 1- n =

=

Po

dT = 3P0V0 – 2 P0V dv

dT 3P0 V0 - 2P0 V =0 = nRV0 dv

\ n=

q

...(ii)

From (i) & (ii)

3Po c

...(i)

nRT v

\p=

3 æ P V ö 5 æ 2P V ö DQ = R ç 0 0 ÷ + R ç 0 0 ÷ 2 è R ø 2 è R ø

24.

- P0 P = V V + 3P 0

[slope =

Þ

23.

P-67

R 1- n

\ C - Cv =

R C - Cv

\ 1-

R 1- n

R =n C - Cv

C - Cv - R C - C v - Cp + Cv = C - Cv C - Cv C - Cp C - Cv

(Q C p - C v = R )

EBD_7764 P-68

Physics

12

Kinetic Theory 1.

2.

3.

4.

5.

6.

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will [2002] (a) increase (b) decrease (c) remain same (d) decrease for some, while increase for others At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C? [2002] (a) 80 K (b) –73 K (c) 3 K (d) 20 K. A gaseous mixture consists of 16 g of helium C and 16 g of oxygen. The ratio p of the mixture Cv is [2005] (a) 1.62 (b) 1.59 (c) 1.54 (d) 1.4 The speed of sound in oxygen (O2) at a certain temperature is 460 ms–1. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) [2008] (a) 1421 ms–1 (b) 500 ms–1 (c) 650 ms–1 (d) 330 ms–1 One kg of a diatomic gas is at a pressure of 8 × 104N/m2. The density of the gas is 4kg/m3. What is the energy of the gas due to its thermal motion? [2009] (a) 5 × 104 J (b) 6 × 104 J (c) 7 × 104 J (d) 3 × 104 J A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g. It is moving with speed v and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by: [2011]

(a) (c) 7.

( g - 1) Mv 2 K 2 gR

( g - 1) Mv 2 K 2R

(b) (d)

gM 2v K 2R ( g - 1) Mv 2 K 2( g + 1) R

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is : Cp ö æ çg = ÷ Cv ø è

[2015]

g +1 g -1 (b) 2 2 3g + 5 3g - 5 (c) (d) 6 6 The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be : [2017]

(a)

8.

(a) 9.

2.5 ´ 10

25

(b)

-2.5 ´ 10

25

(c) -1.61 ´ 1023 (d) 1.38 ´ 1023 Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that [2017] Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is : (a) a = 14 b (b) a = 28 b (c)

a=

1 b 14

(d)

a=b

P-69

Kinetic Theory

1 (c)

1. 2.

2 (d)

3 (a)

4 (a)

5 (a)

6 (c)

Answer Key 7 8 9 (a) (b) (a)

(c) Since P and V are not changing, so temperature remain same. 8RT pM

(d) vrms =

TH 2

\

=

TO2

= For vrms to be equal M M O2 H2

\ 3.

TH 2 2

=

5.

29 R ´ 23 29 R = 9´4 18

Cp =

n1C p1 + n2 C p2 (n1 + n2 )

n1 + n2

and

4.

v=

gRT M

460 = 1421 m / s 0.3237

mass 1 = m3 density 4

1 2 R mv = nCv DT = n DT 2 g -1 1 2 m R mv = DT 2 M g -1

\ DT = 7.

(a)

t=

Cp

47 R 18 = ´ = 1.62 18 29 R Cv (a) The speed of sound in a gas is given by Þ \

0.3237

=

5 5m 5 m PM ´ nRT = RT = 2 2M 2M d [Q PM = dRT ] 5 mP 5 1 ´ 8 ´ 104 = = ´ = 5 ´ 104 J 2 d 2 4 (c) Here, work done is zero. So, loss in kinetic energy = change in internal energy of gas

5R 1 7 R + ´ 2 2 2 1ö æ çè 4 + ÷ø 2

7 10 R + R 4 = 47 R = 9 18 2

vO 2

K.E =

6.

=

M O2

M He g He

5 5 1 PV = ´ 8 ´ 104 ´ = 5 ´ 10 4 J 2 2 4 Alternatively:

n1Cv1 + n2Cv2

4´

´

K.E =

3 1 5 5 4 ´ R + ´ R 6R + R 2 2 2 = 4 = 9 1ö æ çè 4 + ÷ø 2 2 =

g O2

1.4 4 ´ = 0.3237 32 1.67

(a) Volume =

320 Þ TH 2 = 20 K 32

(a) For mixture of gas, Cv =

vHe

=

\ vHe =

Here M H 2 = 2; M O2 = 32 ;

TO 2 = 47 + 273 = 320 K

vO2

tµ

Mv 2 ( g - 1) K 2R

1 æ N ö 3RT 2pd2 ç ÷ èVø M V

T

As, TVg–1 = K So, t µ Vg + 1/2 Therefore, q =

g +1 2

EBD_7764 P-70

8.

Physics

(b) Given: Temperature Ti = 17 + 273 = 290 K

1 ´ 105 ´ 30 1 ö æ 1 ´ 6.023 ´ 10 23 ç ÷ 8.314 è 300 290 ø

Temperature Tf = 27 + 273 = 300 K

=

Atmospheric pressure, P0 = 1 × 105 Pa

= – 2.5 × 1025

Volume of room, V0 = 30 m3 Difference in number of molecules, nf – ni = ?

9.

(a) As we know, Cp – Cv = R where Cp and Cv are molar specific heat capacities R M

Using ideal gas equation, PV = nRT(N0),

or, Cp – Cv =

N0 = Avogadro's number PV Þ n= (N ) RT 0

For hydrogen (M = 2) Cp – Cv = a =

\

nf – ni =

P0V0 æ 1 1 ö - ÷N ç R è T f Ti ø 0

R 2 R For nitrogen (M = 28) Cp – Cv = b = 28 a = 14 or, a = 14b \ b

13

Oscillations 1.

2.

3.

4.

In a simple harmonic oscillator, at the mean position [2002] (a) kinetic energy is minimum, potential energy is maximum (b) both kinetic and potential energies are maximum (c) kinetic energy is maximum, potential energy is minimum (d) both kinetic and potential energies are minimum If a spring has time period T, and is cut into n equal parts, then the time period of each part will be [2002] (a) T n (b) T / n (c) nT (d) T A child swinging on a swing in sitting position, stands up, then the time period if the swing will [2002] (a) increase (b) decrease (c) remains same (d) increases if the child is long and decreases if the child is short A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the

velocities, during oscillation, are equal, the ratio of amplitude of A and B is [2003]

6.

7.

8.

mass is increased by m, the time period m 5T .Then the ratio of is [2003] M 3 3 25 (b) (a) 5 9 16 5 (c) (d) 9 3 Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and k2, respectively. If the maximum

becomes

5.

9.

(a)

k1 k2

(b)

k2 k1

(c)

k2 k1

(d)

k1 k2

The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is [2003] (a) 11% (b) 21% (c) 42% (d) 10% The displacement of a particle varies according to the relation x = 4(cos pt + sin pt). The amplitude of the particle is [2003] (a) – 4 (b) 4 (c) 4 2 (d) 8 A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and total energy (T.E) are measured as a function of displacement x. Which of the following statements is true ? [2003] (a) K.E. is maximum when x = 0 (b) T.E is zero when x = 0 (c) K.E is maximum when x is maximum (d) P.E is maximum when x = 0 The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in air. Neglecting frictional force of water and given that the density of the bob is (4/3) × 1000 kg/m3. Which relationship between t and t0 is true? [2004] (a) t = 2t0 (b) t = t0/2 (c) t = t0 (d) t = 4t0

EBD_7764 P-72

10.

Physics

(a) 11.

12.

13.

T -1 = t1-1 + t 2-1 (b)

T 2 = t12 + t 22

(c) T = t1 + t2 (d) T -2 = t1-2 + t 2-2 The total energy of a particle, executing simple harmonic motion is [2004] 2 (a) independent of x (b) µ x 1/ 2 (c) µ x (d) µ x where x is the displacement from the mean position. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency w0. An external force F(t) proportional to cos wt (w ¹ w 0 ) is applied to the oscillator.. The displacement of the oscillator will be proportional to [2004] 1 1 (a) (b) 2 2 2 2 m (w 0 - w ) m (w 0 + w ) m m (c) 2 2 (d) 2 2 w -w (w + w ) 0

15.

(d) a simple harmonic motion with a period

16.

2

d x dt 2

In forced oscillation of a particle the amplitude is maximum for a frequency w1 of the force while

w1 > w2 when damping is large (b) w1 > w2 (c) w1 = w2 (d) w1 < w 2 Two simple harmonic motions are represented by the equations y1 = 0.1 sin æç100pt + p ö÷ and è 3ø y 2 = 0.1 cos pt . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is [2005] -p p (a) (b) 6 3 p -p (c) (d) 6 3 2 The function sin (wt ) represents [2005] (a) a periodic, but not simple harmonic motion p with a period w

2p w The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would [2005] (a) first decrease and then increase to the original value (b) first increase and then decrease to the original value (c) increase towards a saturation value (d) remain unchanged

17. If a simple harmonic motion is represented by

0

the energy is maximum for a frequency w2 of the force; then [2004] (a) w1 < w2 when damping is small and

14.

(b) a periodic, but not simple harmonic motion 2p with a period w p (c) a simple harmonic motion with a period w

A particle at the end of a spring executes S.H.M with a period t1. while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T then [2004]

(a)

+ ax = 0 , its time period is 2p a

(b)

[2005]

2p a

(d) 2pa (c) 2p a 18. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is [2006] (a) 0.01 s (b) 10 s (c) 0.1 s (d) 100 s 19. Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? [2006] (a)

(b)

1 s 4

1 1 s s (d) 3 12 Two springs, of force constants k1 and k2 are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes [2007]

(c)

20.

1 s 6

P-73

Oscillations

m

k1

21.

k2

(a) 2 f (b) f /2 (c) f /4 (d) 4 f A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is[2007] (a) 2p 2 ma 2 n2 (b) p 2 ma 2 n2 1 2 2 ma n (d) 4p 2 ma 2 n2 4 The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10–2 cos pt metre.The time at which the maximum speed first occurs is[2007] (a) 0.25 s (b) 0.5 s (c) 0.75 s (d) 0.125 s A point mass oscillates along the x-axis according to the law x = x0 cos(wt - p / 4) . If the acceleration of the particle is written as a = A cos(wt + d ) ,then [2007]

(c)

22.

23.

(a)

2

A = x0 w , d = 3p / 4

(b) A = x0, d = -p / 4 (c)

A = x0 w 2 , d = p / 4 2

24.

25.

(d) A = x0 w , d = -p / 4 If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?[2009] (a) aT/x (b) aT + 2pv (c) aT/v (d) a2T2 + 4p2v2 Two particles are executing simple harmonic motion of the same amplitude A and frequency w along the x-axis. Their mean position is separated by distance X0(X0 > A). If the maximum separation between them is (X0 + A), the phase difference between their motion is: [2011] (a) (c)

26.

p 3 p 6

(b) (d)

p 4 p 2

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio

æ A1 ö

of çè A ÷ø is: 2

[2011] 1

(a)

M +m M

(c)

æ M + mö 2 çè ÷ M ø

(b)

æ M ö2 çè ÷ M + mø

(d)

M M +m

1

27. A wooden cube (density of wood ‘d’) of side ‘ l ’ floats in a liquid of density ‘r’ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period ‘T’ [2011 RS] (a)

2p

(c)

2p

ld rg ld

(r - d ) g

(b)

2p

(d)

2p

lr dg lr

(r - d ) g

28. If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0s to t = t s, then t may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with b as the constant of proportionality, the average life time of the pendulum in second is (assuming damping is small) [2012] 0.693 (a) (b) b b 1 2 (d) b b 29. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs S1 and S2 of force constants k1 and k2 respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2. Statement 1 : If stretched by the same amount work done on S1 Statement 2 : k1 < k2 [2012] (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is false. (c) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1 (d) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

(c)

EBD_7764 P-74

30.

31.

32.

33.

Physics

(a)

1 AgP0 2p V0 M

(c)

1 2p

A 2 gP0 MV0

(b)

1 V0 MP0 2p A 2 g

(d)

1 2p

E

é æ T é æ T ö2 ù A ê1 - ç M ÷ ú (b) ê1 - ç T êë è M ë è T ø û Mg

d

PE

E PE (b)

KE

E

KE PE

(c)

E

d

PE KE

(d)

d

35. A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant 2A from equilibrium 3 position. The new amplitude of the motion is : [2016]

that it is at a distance

2 ö ù A ÷ ú Mg ø úû

éæ T ö2 ù A éæ T ö2 ù Mg M (c) êç ÷ - 1ú (d) êç M ÷ - 1ú ëè T ø û Mg ëè T ø û A For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one

KE

(a)

MV0 AgP0

A particle moves with simple harmonic motion in a straight line. In first t s, after starting from rest it travels a distance a, and in next t s it travels 2a, in same direction, then: [2014] (a) amplitude of motion is 3a (b) time period of oscillations is 8t (c) amplitude of motion is 4a (d) time period of oscillations is 6t A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young's modulus 1 of the material of the wire is Y then is equal Y to : (g = gravitational acceleration) [2015] (a)

34.

of the following represents these correctly? (graphs are schematic and not drawn to scale) [2015]

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to a times its original magnitude, where a equals [2013] (a) 0.7 (b) 0.81 (c) 0.729 (d) 0.6 An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency [2013]

(a)

(b)

7A 3

A 41 (d) 3A 3 A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like: [2017]

(c) 36.

A 3

P-75

Oscillations

(c)

(a)

(d)

(b)

1 (c) 16 (b) 31 (c)

1.

2.

3.

2 (b) 17 (a) 32 (d)

3 (b) 18 (a) 33 (c)

4 (c) 19 (a) 34 (d)

5 (c) 20 (a) 35 (b)

6 (d) 21 (b) 36 (b)

Answer Key 7 8 9 (c) (a) (a) 22 23 24 (b) (a) (a)

(c) The kinetic energy (K. E.) and potential energy (U) of a simple harmonic oscillator is given by, 1 1 2 2 2 K.E = k ( A - x ) ; U = kx 2 2 Where A = amplitude and k = mw2 x = displacement from the mean position At the mean position x = 0 1 2 \ K.E. = kA = Maximum and U = 0 2 (b) Let the spring constant of the original

m spring be k. Then its time period T = 2p k where m is the mass of oscillating body. When the spring is cut into n equal parts, the spring constant of one part becomes nk. Therefore the new time period, m T T ' = 2p = nk n l (b) The time period T = 2p where l = g distance between the point of suspension and the centre of mass of the child. As shown in the figure, l ¢ < l \ T ¢ < T i.e., the period decreases.

10 (b) 25 (a)

11 (a) 26 (c)

12 (b) 27 (a)

13 (c) 28 (d)

14 (b) 29 (b)

15 (c) 30 (c)

point of suspension l'

l

Case (ii) child standing 4.

(c)

T = 2p

M k

T ' = 2p

M + m 5T = k 3

M +m 5 M = ´ 2p 3 k k 25 M +m = ´M 9 16 m 25 m 25 1+ = Þ = -1 = 9 9 9 M M (c) Maximum velocity during SHM = Aw But k = mw2 k \ w= m \ 2p

5.

Case (i) child sitting

EBD_7764 P-76

Physics

k m Here the maximum velocity is same and m is also same \ Maximum velocity = A

\ A1 k1 = A2 k 2

6.

(d)

l T = 2p g

\

A1 A2

=

k2

\ t = 2p t = 2t0

10. (b) For first spring, t1 = 2p

k1

m k2 when springs are in series then, k1k 2 keff = kl + k 2

=

T '- T ´ 100 T

1.21l - l

´ 100 = l = (1.1 - 1) ´ 100 = 10%

7.

(c)

(

)

x = 4 2 sin(p t + 45°) on comparing it with x = A sin(wt + f)

9.

we get A = 4 2 1 (a) K.E. = mw 2 (a 2 - x 2 ) 2 When x = 0, K.E is maximum and is equal to 1 mw 2 a 2 . 2 l l (a) t = 2p ; t 0 = 2p g g eff Buoyant force 1000 Vg

Weight

4 ´ 1000 Vg 3

æ4 ö 1000 Vg Net force = çè - 1÷ø ´ 1000 Vg = 3 3 g eff =

g 1000 Vg = 4 4 3 ´ ´ 1000 V 3

k1k 2 2

2

2

m m t2 t + = 2p + 1 2 2 k 2 k1 (2p) (2p)

\ T = 2p

x = 4(cos pt + sin pt )

= 4 2(sin p t cos 45° + cos p t sin 45°)

m ( kl + k 2 )

\ T = 2p

1.21 - 1 ´ 100

æ sin pt cos pt ö = 2 ´ 4ç + ÷ è 2 2 ø

8.

m , k1

For second spring, t 2 = 2p

1.21l and T ' = 2p g

(Q l ' = l + 21% of l) % increase =

l g/4

2

2

Þ T = t1 + t 2 where x is the displacement from the mean position 11. (a) At any instant the total energy is 1 2 kA = constant, where A0 = amplitude 2 0 hence total energy is independent of x. 12. (b) Equation of displacement is given by x = A sin(wt + f) F0

where A =

2

2 2

m (w 0 - w ) F0

=

m (w 0 2 - w 2 ) Here damping effect is considered to be zero

\x µ

1 2

2

m (w 0 - w ) 13. (c) Since energy µ ( Amplitude) 2 , the maximum for both of them occurs at the same frequency \ w1 = w2

14. (b)

v1 =

dy1 pö æ = 0.1 ´ 100p cos ç100pt + ÷ è dt 3ø

v2 =

dy2 pö æ = - 0.1p sin pt = 0.1p cos ç pt + ÷ è dt 2ø

Oscillations

\ Phase diff. = f1 - f 2 =

p p 2 p - 3p - = 3 2 6

p 6 (c) Clearly sin2 wt is a periodic function as sin p wt is periodic with period w

=–

15.

P-77 19. (a) K.E. of a body undergoing SHM is given by,

K .E. =

1 2 2 2 ma w cos wt 2

1 2 2 ma w 2 Given K.E. = 0.75 T.E. T .E. =

2

Þ 0.75 = cos wt Þ wt =

p 6

p p´2 1 Þt= Þt= s 6´w 6 ´ 2p 6 20. (a) The two springs are in parallel. \ Effective spring constant, k = k1 + k2 Now, frequency of oscillation is given by Þt=

0

p/w 2

For SHM

d y dt

2

2p/w

3p/w

µ -y

dy = 2w sin wt cos wt = w sin 2wt dt 2

d y

16.

2

= 2w cos 2wt which is not dt 2 proportional to –y. Hence, it is not in SHM. (b) Centre of mass of combination of liquid and hollow portion (at position l ), first goes down ( to l + D l) and when total water is drained out, centre of mass regain its original position (to l ),

l g \ ‘T ’ first increases and then decreases to original value. T = 2p

c 2

17.

(a)

d x dt

2

2 = -ax = -w x

Þ w = a or T = 18.

2p 2p = w a

(a) Maximum velocity, vmax = a w 2p vmax = a ´ T -3 2pa 2 ´ 3.14 ´ 7 ´ 10 ÞT = = » 0.01 s vmax 4.4

f =

1 2p

k m

1 k1 + k 2 ....(i) 2p m When both k1 and k2 are made four times their original values, the new frequency is given by

or, f =

f '=

1 2p

4 k1 + 4k 2 m

æ 1 k1 + k 2 ö 4(k1 + 4k 2 ) = 2ç ÷ m m ø è 2p 21. (b) The kinetic energy of a particle executing S.H.M. is given by 1 K = ma2 w2 sin2wt 2 where, m = mass of particle a = amplitude w = angular frequency t = time 1 Now, average K.E. = < K > = < mw2 a2 2 sin2 wt > 1 = mw2a2 2 1ö æ 1ö æ 1 2 = mw2a2 çè ÷ø çè Q < sin q > = ÷ø 2 2 2 1 1 2 2 = mw a = ma2 (2pn) 2 (Q w = 2pn) 4 4 =

1 2p

2

2 2

or, < K > = p ma n

EBD_7764 P-78

22.

23.

24.

25.

Physics

x = 2 × 10–2 cos p

(b) Here, t Speed is given by dx v= = 2 × 10–2 p sin p t dt For the first time, the speed to be maximum, p sin p t = 1 or, sin p t = sin 2 1 p Þ pt = or,, t = = 0.5 sec. 2 2 (a) Here, x = x0 cos (wt – p / 4 ) dx pö æ = - x0 w sin ç wt - ÷ \ Velocity, v = è dt 4ø Acceleration, dv pö æ 2 a= = - x0w cos ç wt - ÷ è dt 4ø é p ù æ ö 2 = x0w cos ê p + ç wt - ÷ ú 4øû ë è 3 p æ ö = x0 w 2 cos ç wt + ÷ ...(1) è 4ø Acceleration, a = A cos (wt + d) ...(2) Comparing the two equations, we get 3p A = x0w2 and d = . 4 (a) For an SHM, the acceleration a = -w2 x a 2 where w is a constant. Therefore, is a x constant. The time period T is also aT constant. Therefore, is a constant. x2 (a) Let, x1 = A sin w t and x = A sin (w t + f)

27. (a)

d

At equilibrium Fb = mg rAl 0 g = dAlg

l0 + x

Restoring force, F = mg – Fb'

F = mg - rA ( l 0 + x ) g

dAla = dAlg - rAl 0 g - rgAx

rg x dl Therefore, wooden cube performs S.H.M. a=-

26.

p

Þ f= 3 (c) The net force becomes zero atthe mean point. Therefore, linear momentum must be conserved. \ Mv1 = (M + m)v2 MA1

k k = ( M + m) A2 M m+M

æ

k ö

\ çV = A M ÷ è ø Q A1 M = A2 M + m A1 \ A = 2

m+ M M

....(i)

d

As the maximum separation between the particles is A, f 2

r

l0

fö f æ x2 – x1 = 2A cos çè wt + 2 ÷ø sin 2

\ 2 A sin = A

l

rg ld Þ T = 2p dl rg (d) The equation of motion for the pendulum, suffering retardation F = – kx - bv

\ w=

28.

2

Þ m d x + kx + b dk = 0 dt dt 2 2 d x k b dx + x+ =0 Þ m dt dt 2 m Þ

d2 x dt

2

+

b dx k + x =0 m dt m

… (1)

P-79

Oscillations

Let x = e

lt

is the solution of the equation (1)

g

P=

2

dx d x = le l t Þ = l 2 e lt 2 dt dt Substituting in the equation (1) b k l 2 e lt + l e lt + e l t = 0 m m b k l2 + l + = 0 m m

b b2 k ± -4 2 -b ± b 2 - 4km m m m l= = 2 2m Solving the equation (1) for x, we have -

-b

x = e 2m

t

k +b w = w0 - l where w0 = , l = m 2 1 2 The average life = = l b 2

29.

2

1 1 2 k1x 2 ; w2 = k2 x 2 2 Since w1 > w2 Thus (k1 > k2)

w1 =

30.

bt 2m

(c) Q A = A 0e (where, A0 = maximum amplitude) According to the questions, after 5 second, -

b(5) 2m

… (i) 0.9A 0 = A 0e After 10 more second, -

b(15)

…(ii) A = A0 e 2m From eqns (i) and (ii) A = 0.729 A0 \ a = 0.729 31.

(c)

Piston x x0

Cylinder containing ideal gas

Let piston is displaced by distance x æ P xg ö 0 0 Mg - ç ÷ A = Frestoring çè ( x - x )g ÷ø 0 æ ö x0g P0 A ç1 ÷ = Frestoring çè ( x - x )g ÷ø 0

[ x0 - x » x0 ] gP0 Ax x0 \ Frequency with which piston executes SHM.

P0V0 g = PV g Mg = P0A P0 Ax0 = PA( x0 - x)

f =

gP0 A 1 x0 M = 2p

gP0 A2 MV0

32. (d) In simple harmonic motion, starting from rest, At t = 0 , x = A x = Acoswt ..... (i) When t = t , x = A – a When t = 2 t , x = A –3a From equation (i) A – a = Acosw t ......(ii) A – 3a = A cos2w t ....(iii) As cos2w t = 2 cos2 w t – 1...(iv) From equation (ii), (iii) and (iv) 2

A - 3a 2 A2 + 2a 2 - 4 Aa - A2 = A A2 2 2 2 Þ A – 3aA = A + 2a – 4Aa Þ 2a2 = aA Þ A = 2a Þ

… (1) g

1 2p

A - 3a æ A-a ö = 2ç ÷ -1 A è A ø

Mg = P0 A

g

( x0 - x )g

F=-

1 2 (b) Q w = kx 2

-

P0 x0

EBD_7764 P-80

Physics

a 1 = A 2 Now, A – a = A coswt

Þ

Þ

cos wt =

A- a A

Þ

cos wt =

1 2

Þ T= 6t 33.

or

Initially

2 æ 2A ö V= w A -ç ÷ è 3 ø

2

2 æ 2A ö 3v = w A' - ç ÷ è 3 ø Where A'= final amplitude (Given at x = 2A , velocity to trebled) 3

Finally 2p p t= T 3

l g When additional mass M is added then

(c) As we know, time period, T = 2p

On dividing we get

l + Dl TM = 2p g TM T

2

é Mgl ù êQ Dl = Ay ú ë û 2 ù 1 éæ T ö A \ = êç M ÷ - 1ú y ëêè T ø Mg ûú

(d) K.E =

1 k ( A2 - d 2 ) 2

1 2 kd 2 At mean position d = 0. At extrement positions d = A

and P.E. =

35.

2 2 (b) We know that V = w A - x

3 = 1

æ 2A ö A' 2 - ç ÷ è 3 ø æ 2A ö A2 - ç ÷ è 3 ø

2

2

é 2 4A 2 ù 4A 2 9 ê A - 9 ú = A'2 – 9 ëê ûú

2 l + Dl l + Dl æ TM ö = = or ç ÷ l è T ø l

Mg æ TM ö =1+ or, ç è T ÷ø Ay

34.

2

7A 3 36. (b) For a particle executing SHM At mean position; t = 0, wt = 0, y = 0, V = Vmax = aw

\ A' =

1 mw2a2 2 T p At extreme position : t = , wt = , y = A, V = 4 2 Vmin = 0 \ K.E. = KEmin = 0 1 Kinetic energy in SHM, KE = mw2(a2 – y2) 2 1 = mw2a2cos2wt 2 Hence graph (2) correctly depicts kinetic energy time graph.

\

K.E. = KEmax =

P-81

Waves

14

waves 1.

2.

3.

4.

5.

6.

Length of a string tied to two rigid supports is 40 cm. Maximum length (wavelength in cm) of a stationary wave produced on it is [2002] (a) 20 (b) 80 (c) 40 (d) 120 Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is [2002] (a) 1 : 2 (b) 1 : 4 (c) 2 : 1 (d) 4 : 1. A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is [2002] (a) 286 cps (b) 292 cps (c) 294 cps (d) 288 cps A wave y = a sin(wt–kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is [2002] (a) y = a sin( w t + kx) (b) y = –a sin( w t + kx) (c) y = a sin( w t – kx) (d) y = –a sin( w t – kx) When temperature increases, the frequency of a tuning fork [2002] (a) increases (b) decreases (c) remains same (d) increases or decreases depending on the material The displacement y of a wave travelling in the x -direction is given by

7.

8.

9.

pö æ y = 10 - 4 sinç 600t - 2x + ÷ metres 3ø è where x is expressed in metres and t in seconds. The speed of the wave - motion, in ms–1 , is [2003] (a) 300 (b) 600 (c) 1200 (d) 200 A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg-wt between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is [2003] (a) 50 Hz (b) 100 Hz (c) 200Hz (d) 25Hz A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was [2003] (a) (256 + 2) Hz (b) (256 – 2) Hz (c) (256 – 5) Hz (d) (256 + 5) Hz The displacement y of a particle in a medium can be expressed as,

pö æ y = 10-6 sin ç100t + 20 x + ÷ m where t is in è 4ø second and x in meter. The speed of the wave is [2004] (a) 20 m/s (b) 5 m/s (c) 2000 m/s (d) 5p m/s

EBD_7764 P-82

10.

11.

12.

13.

14.

15.

Physics

When two tuning forks ar e sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2? [2005] (a) 202 Hz (b) 200 Hz (c) 204 Hz (d) 196 Hz An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ? [2005] (a) 0.5% (b) zero (c) 20 % (d) 5 % A whistle producing sound waves of frequencies 9500 HZ and above is approaching a stationary person with speed v ms–1 . The velocity of sound in air is 300 ms–1. If the person can hear frequencies upto a maximum of 10,000 HZ, the maximum value of v upto which he can hear whistle is [2006] 15 ms -1 (a) 15 2 ms -1 (b) 2 (c) 15 ms -1 (d) 30 ms-1 A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is [2006] (a) 105 Hz (b) 1.05 Hz (c) 1050 Hz (d) 10.5 Hz A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of [2007] (a) 100 (b) 1000 (c) 10000 (d) 10 While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she

16.

17.

18.

19.

measures the column length to be x cm for the second resonance. Then [2008] (a) 18 > x (b) x > 54 (c) 54 > x > 36 (d) 36 > x > 18 A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos (a x – bt). If the wavelength and the time period of the wave are 0.08 m and 2.0s, respectively, then a and b in appropriate units are [2008] (a) a = 25.00 p , b = p 0.08 2.0 ,b = (b) a = p p 0.04 1.0 ,b = (c) a = p p p (d) a = 12.50p, b = 2.0 Three sound waves of equal amplitudes have frequencies (n –1), n, (n + 1). They superpose to give beats. The number of beats produced per second will be : [2009] (a) 3 (b) 2 (c) 1 (d) 4 A motor cycle starts from rest and accelerates along a straight path at 2m/s2. At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (Speed of sound = 330 ms–1) [2009] (a) 98 m (b) 147 m (c) 196 m (d) 49 m The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by

é æ t x öù y = 0.02(m) sin ê2p ç ÷ú . ë è 0.04(s ) 0.50(m) ø û The tension in the string is [2010] (a) 4.0 N (b) 12.5 N (c) 0.5 N (d) 6.25 N 20. The transverse displacement y (x, t) of a wave

on a string is given by y( x, t ) = e

(

- ax 2 + bt 2 + 2 ab ) xt

).

This represents a: [2011] (a) wave moving in – x direction with speed b a

Waves

(b) standing wave of frequency (c) standing wave of frequency

b 1 b

(d) wave moving in + x direction speed 21.

22.

23.

1 (b) 16 (a)

a b

A travelling wave represented by y = A sin (wt – kx) is superimposed on another wave represented by y = A sin (wt + kx). The resultant is [2011 RS] (a) A wave travelling along + x direction (b) A wave travelling along – x direction (c) A standing wave having nodes at nl x= , n = 0,1, 2.... 2 (d) A standing wave having nodes at 1ö l æ x = ç n + ÷ ; n = 0,1, 2.... è 2ø 2 Statement - 1 : Two longitudinal waves given by equations : y1 ( x, t ) = 2a sin(wt - kx ) and y2 ( x, t ) = a sin (2wt - 2kx) will have equal intensity. [2011 RS] Statement - 2 : Intensity of waves of given frequency in same medium is proportional to square of amplitude only. (a) Statement-1 is true, statement-2 is false. (b) Statement-1 is true, statement-2 is true, statement-2 is the correct explanation of statement-1 (c) Statement-1 is true, statement-2 is true, statement-2 is not the correct explanation of statement-1 (d) Statement-1 is false, statement-2 is true. A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the aircolumn is now : [2012] (a) f (b) f / 2 (c) 3 f /4 (d) 2 f

2 (c) 17 (b)

3 (b) 18 (a)

4 (b) 19 (d)

5 (b) 20 (a)

6 (a) 21 (d)

P-83 24. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively? [2013] (a) 188.5 Hz (b) 178.2 Hz (c) 200.5 Hz (d) 770 Hz 25. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. [2014] (a) 12 (b) 8 (c) 6 (d) 4 26. A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms–1) close to : [2015] (a) 18% (b) 24% (c) 6% (d) 12% 27. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now : [2016] (a) 2f (b) f

f 3f (d) 2 4 28. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is : [2016] –2 (take g = 10 ms )

(c)

(a)

2 2s

(b)

(c)

2p 2 s

(d) 2 s

Answer Key 7 8 9 (a) (c) (b) 22 23 24 (a) (a) (b)

10 (d) 25 (c)

11 (c) 26 (d)

12 (c) 27 (b)

2s

13 (a) 28 (a)

14 (a)

15 (b)

EBD_7764 P-84

1.

Physics

(b) This will happen for fundamental mode of vibration as shown in the figure. Here, S1

2.

4.

l = 40 \ l = 80 cm 2

S2

40 cm S1 and S2 are rigid support (c) The fundamental frequency for closed organ pipe is given by

5.

lù v é êëQl = 4 úû 4l Where l = length of the tube and v is the velocity of sound in air. uc =

and the frequency of known fork. This would produce high beat frequency. (b) To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of p. The equation of the reflected wave will be y = a sin (wt + kx + p) Þ y = – a sin (wt + kx) (b) The frequency of a tuning fork is given by the expression f=

u0 =

v 2l

4 3 pl

2

Y r

As temperature increases, l increases and therefore f decreases.

l = l /4

The fundamental frequency for open organ pipe is given by

m2 k

6.

(a)

pö æ y = 10-4 sin ç 600t - 2x + ÷ è 3ø But y = A sin ( wt - kx + f)

lù é êëQl = 2 úû

On comparing we get w = 600; k=2 Also velocity of wave is given by

l l= 2

w 600 = = 300 ms -1 k 2 (a) For a string vibrating between two rigid support, the fundamental frequency is given by v=

u0 v 4l 2 \ u = 2l ´ v = 1 c

3.

(b) A tuning fork produces 4 beats/sec with another tuning fork of frequency 288 cps. From this information we can conclude that the frequency of unknown fork is 288 + 4 cps or 288 – 4 cps i.e. 292 cps or 284 cps. When a little wax is placed on the unknown fork, it produces 2 beats/sec. When a little wax is placed on the unknown fork, its frequency decreases and simultaneously the beat frequency decreases confirming that the frequency of the unknown fork is 292 cps. NOTE Had the frequency of unknown

fork been 284 cps, then on placing wax its frequency would have decreased thereby increasing the gap between its frequency

7.

n=

v 1 T = 2l 2l m

l=

=

l 2

1 10 ´ 9.8 = 50 Hz 2 ´ 1 9.8 ´ 10-3

As the string is vibrating in resonance to a.c of frequency n, therefore both the frequencies are same.

Waves 8. (c) A tuning fork of frequency 256 Hz makes 5 beats/second with the vibrating string of a piano. Therefore, the frequency of the vibrating string of piano is (256 ± 5) Hz. i.e., either 261Hz or 251 Hz. When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is 251Hz

9.

10.

11.

12.

P-85

n=1 Therefore lowest resonant frequency = 105 Hz. æI ö 14. (a) We have, L1 = 10 log ç 1 ÷ ; è I0 ø

æI ö L2 = 10 log ç 2 ÷ è I0 ø

(b) From equation given, w = 100 and k = 20 w 100 = 5m / s v= = k 20 (d) Frequency of fork 1 = 200 Hz = n 0 No. of beats heard when fork 2 is sounded with fork 1 = Dn = 4 Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by, n = n0 – Dn = 200 – 4 = 196 Hz vù é êv + 5 ú é v + v0 ù é6ù = nê (c) n ' = n ê ú = nê ú ú ë5û ë v û ê v ú úû êë n' 6 n' - n 6 - 5 = ; = ´ 100 = 20% n 5 n 5 é v ù (c) n' = n ê ú ë v - vs û

æI ö æI ö \ L1 – L2 = 10 log ç 1 ÷ - 10log ç 2 ÷ è I0 ø è I0 ø æI I ö or, DL = 10 log ç 1 ´ 0 ÷ è I0 I 2 ø

æI ö or, DL = 10log ç 1 ÷ è I2 ø æI ö æI ö or, 20 = 10 log ç 1 ÷ or, 2 = log ç 1 ÷ è I2 ø è I2 ø

or,

Þ Intensity decreases by a factor 100. 15. (b) For first resonant length

For second resonant length n' = \

Þ 300 - v = 300 ´ 0.95 Þ v = 300 - 285 = 15 ms -1

v v = 315 Þ = 105 Hz 2l 2l The lowest resonant frequency is when

v 3v' = 4 ×18 4× x v' v

v' cm v v' > v because velocity of light is greater in \ x = 54 ´

n + 1 420 = Þn =3 n 315

Hence 3 ´

3v ' 3v ' (in summer) = 4l 2 4x

\ x = 3 ´ 18 ´

v = 420 (a) Given nv = 315 and (n + 1) 2 l 2l Þ

v v (in winter) = 4l1 4 ´ 18

n=

é 300 ù Þ 10000 = 9500 ê ú ë 300 - v û

13.

I1 I = 102 or, I2 = 1 . I2 100

summer as compared to winter (v µ T )

\ x > 54cm 16

(a)

y(x, t) = 0.005 cos (ax - bt) (Given) Comparing it with the standard equation of wave

EBD_7764 P-86

Physics

Velocity on a string is given by

y(x, t) = a cos (kx - wt) we get k = a and w = b 2p 2p and w = l T 2p 2p Þ = a and =b l T Given that l = 0.08 m and T = 2.0s 2p 2p =p \ a= = 25p and b = 2 0.08 (b) Maximum number of beats

But k =

17.

20.

\ T = v2 ´ m = (12.5)2 × 0.04 = 6.25 N (a) Given wave equation is y(x,t)

= e(- ax

= ( n + 1) – ( n – 1) = 2 18.

(a)

2

u=0 Electric siren

s

21.

(d)

é æ t ö x ù y = 0.02(m)sin ê2p ç ú ÷ ë è 0.04( s) ø 0.50(m) û Comparing this equation with the standard wave equation y = a sin(wt - kx ) we get

w=

2p 0.04

Þ n=

1 = 25 Hz 0.04

2p Þ l = 0.5 m 0.50 \ velocity, v = nl = 25 × 0.5 m/s = 12.5 m/s k=

ax )2 + ( b t ) 2 + 2 a x . b t ]

-( =e

ax + bt ) 2 2

(d)

b a

y = A sin (wt – kx) + A sin (wt + kx) For standing wave nodes cos kx = 0 2p p .x = (2n + 1) l 2

é v - vm ù n ' =n ê ú ë v û

19.

-[( =e

y = 2A sin wt cos kx

\ vm = 2 s According to Doppler’s effect

Þ s = 98.01 m

)

Þ Speed of wave =

\ v2m = 2 ´ 2 ´ s

é 330 - 2 s ù 0.94n = n ê ú ë 330 û

+bt 2 + 2 ab xt

= It is a function of type y = f (x + vt)

Motor cycle

v 2m - u 2 = 2as

2

æ ö b ÷ - çè x + tø a e

vm

a = 2m/s

T m

v=

\ x=

22.

( 2n + 1) l , n = 0,1, 2,3,........... 4

(a) Since, I µ A2w 2 I1 µ (2a)2 w 2 I 2 µ a 2 (2w ) 2

23.

I1 = I 2 Intensity depends on frequency also. (a) The fundamental frequency of open tube v … (i) + n0 = 2l 0 where l is the length of the tube v = speed of sound That of closed tube v nc = … (ii) 4lc l0 According to the problem lc = 2 v v Þ nc = Thus nc = … (iii) l0 / 2 2l

P-87

Waves

From equations (i) and (iii) n 0 = nc 24.

æ f2 ö æ 300 ö çè f - 1÷ø ´ 100 = çè 340 - 1÷ø ´ 100 ; 12% 1

Thus, nc = f ( Q n0 = f is given) (b) Fundamental frequency,

f =

v 1 T 1 = = 2l 2l m 2l

27.

T Ar

l

é T mù and m = ú êQ v = m lû ë

Also, Y =

Þ f =

Dl , r and g in eqn. l

(i) we get, 2 103 f = ´ 7 3 or f » 178.2 Hz 25.

v 2l

The fundamental frequency in case (b) is

....(i)

Dl = 0.01, l r = 7.7 × 103 kg/m3 (given) g = 2.2 × 1011 N/m2 (given)

(c) Length of pipe = 85 cm = 0.85m Frequency of oscillations of air column in closed organ pipe is given by,

(b)

The fundamental frequency in case (a) is

f=

l = 1.5 m,

Putting the value of l,

l (a)

Tl T Y Dl Þ = ADl A l

1 gDl 2l lr

f

(b)

f'= 28.

(a)

We know that velocity in string is given by

where m =

Þ 26.

(d)

...(i)

m mass of string = l length of string

The tension T = From (1) and (2)

m ´ x´g l

l T x

(2n - 1)u f = £ 1250 4L

Þ

T m

v=

(2n - 1)u f = 4L

(2n - 1) ´ 340 £ 1250 0.85 ´ 4 2n – 1 < 12.5 » 6

v u = =f 4(l / 2) 2l

dx = gx dt

x -1/2 dx = g dt l

l

é v ù 320 f1 = f ê Hz ú= f´ v v 300 sû ë

\

é v ù 320 f2 = f ê Hz ú= f ´ + v v 340 sû ë

= g´t \ t = 2

òx 0

-1/2

dx - g ò dt 2 l 0

20 l =2 =2 2 g 10

..(ii)

EBD_7764 P-88

Physics

15

Electric Charges and Fields 1.

A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is [2002] E

F

D

c

O H A

2.

3.

4.

5.

q

q G B

L

(a) q /4 p Î0 L

(b) zero

(c) q/2 p Î0 L

(d) q/3 p Î0 L

If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is [2002] (a) Q/2 (b) –Q/2 (c) Q/4 (d) –Q/4 If the electric flux entering and leaving an enclosed surface respectively is f1 and f 2 , the electric charge inside the surface will be [2003] (a) (f2 - f1 )e o (b) (f1 + f2 ) / e o (c) (f2 - f1) / e o (d) (f1 + f2 )e o Three charges –q1 , +q2 and –q3 are place as shown in the figure. The x - component of the force on –q1 is proportional to [2003]

6.

7.

Y

-q 3

8. a

b +q 2 X

-q 1

(a)

q2

(c)

q2

b

2

b2

+

q3 2

a q3

a2

cos q cos q

(b)

q2

(d)

q2

b

2

b2

+ -

q3 a2 q3 a2

sin q sin q

Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is [2004] (a) F/8 (b) 3 F/4 (c) F/4 (d) 3 F/8 Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [2004] (a)

-

Q (1 + 2 2) 2

(b)

Q (1 + 2 2) 4

(c)

-

Q (1 + 2 2) 4

(d)

Q (1 + 2 2) 2

A charged oil drop is suspended in a uniform field of 3×104 v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9×10–15 kg and g = 10 m/s2) [2004] (a) 1.6×10–18 C (b) 3.2×10–18 C (c) 3.3×10–18 C (d) 4.8×10–18 C Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is [2005] (a)

L 4

(c) 4 L

(b) 2 L (d) 8 L

P-89

Electric Charges and Fields

9.

A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface charge density s of the sheet is proportional to [2005]

E(r)

E(r)

(a)

(b) O

r

R

E(r)

P

q S

10.

11.

12.

13.

B (a) cot q (b) cos q (c) tan q (d) sin q An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience [2006] (a) a translational force only in the direction of the field (b) a translational force only in a direction normal to the direction of the field (c) a torque as well as a translational force (d) a torque only Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is [2006] (a) 4 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4 If gE and gM are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio [2007] electronic charge on the moon to be electronic charge on the earth (a) g M / g E (b) 1 (c) 0 (d) g E / g M A thin spherical shell of radus R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 £ r < ¥, where r is the distance from the centre of the shell? [2008]

(c)

r

R

O

E(r)

r

R

O

(d)

r

R

O

14. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals: [2009] (a) –1 (b) 1 (c)

1 2

-

(d)

15. Let r (r ) =

Q

p R4

-2 2

r be th e charge density

distribution for a solid sphere of radius R and total charge Q. For a point ‘P’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is : [2009] (a) (c)

Q 4p

(b)

Î0 r12

Qr12

Qr12 4p Î0 R4

(d) 0

3p Î0 R4

16. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net ur field E at the centre O is [2010] j

(a)

i

O

q 2

4p e 0 r

2

ˆj

q

(b)

-

q 2

4p e 0 r 2

q

ˆj

ˆj 2p e 0 r 2 2p e 0 r 17. Let there be a spherically symmetric charge distribution with charge density varying as æ5 rö r(r ) = r0 ç - ÷ upto r = R , and r(r ) = 0 è 4 Rø (c)

-

2

2

ˆj

(d)

2

EBD_7764 P-90

Physics

for r > R , where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by [2010] r0 r æ 5 r ö 4pr0 r æ 5 r ö (a) ç - ÷ (b) 3e çè 3 - R ÷ø 4e 0 è 3 R ø 0

r0 r æ 5 r ö r0 r æ 5 r ö ç - ÷ (d) 3ε çè 4 - R ÷ø 4ε0 è 4 R ø 0 Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d 0 per unit area. The disc rotates about its axis with a uniform angular speed w.The magnetic moment of the disc is [2011 RS] pR 4 sw

(b)

pR 4 sw 2

pR 4 sw (d) 2pR 4 sw 4 Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp, rd and ra. Which one of the following relation is correct? [2012]

(c)

ra = rp < rd

rd > rp (c) ra > rd > rp (d) ra = 30. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is [2013] (a) 9.1 × 10–11 weber (b) 6 × 10–11 weber (c) 3.3 × 10–11 weber (d) 6.6 × 10–9 weber 31. A conductor lies along the z-axis at -1.5 £ z < 1.5 m and carries a fixed current of 10.0 A in -aˆ z direction (see figure). For a field r B = 3.0 ´10-4 e -0.2x aˆ y T, fin d the power required to move the conductor at constant

An electric charge +q moves with velocity r v = 3iˆ + 4 ˆj + kˆ in an electromagnetic field ur ur given by E = 3i$ + $j + 2k$ and B = iˆ + ˆj - 3kˆ

(a)

29.

d

ra = rp = rd

z 1.5

I

y

B

2.0 x

–1.5

(a) 1.57 W (b) 2.97 W (c) 14.85 W (d) 29.7 W 32. A rectangular loop of sides 10 cm and 5 cm carrying a current 1 of 12 A is placed in different orientations as shown in the figures below : [2015] z

I

B

I

I

(A)

x

I

y

P-121 uur uur (a) F1 is radially inwards and F2 = 0 uur uur (b) F1 is radially outwards and F2 = 0 uur uur (c) F1 = F2 = 0 uur uur (d) F1 is radially inwards and F2 is radially outwards 34. Two identical wires A and B, each of length 'l', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at the centres of the circle and BA square respectively, then the ratio is: BB [2016]

Moving Charges and Magnetism

z B (B)

I

I I

x

y

I

z B

I (C)

I

I

y

I

x

z B (D)

I

I

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ? (a) (B) and (D), respectively (b) (B) and (C), respectively (c) (A) and (B), respectively (d) (A) and (C), respectively Two coaxial solenoids of different radius carry uur current I in the same direction. F1 be the

outer solenoid due to the inner one. Then : [2015]

2 (a) 17 (d) 32 (a)

3 (a) 18 (d) 33 (c)

4 (c) 19 (b) 34 (b)

5 (a) 20 (b) 35 (c)

6 (b) 21 (c) 36 (c)

(b)

p2 8 2

p2 p (d) 8 16 2 35. A galvanometer having a coil resistance of 100 W gives a full scale deflection, when a currect of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is : [2016] (a) 0.1 W (b) 3W (c) 0.01W (d) 2W 36. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15W, it shows full scale deflection. The value of the resistance to be put in series with th e galvanometer to convert it into to voltmeter of range 0 - 10 V is [2017] (a) 2.535× 103 W (b) 4.005 × 103 W (c) 1.985 × 103 W (d) 2.045 × 103 W

(c)

magnetic force on the inner solenoid due to the uur outer one and F2 be the magnetic force on the

1 (b) 16 (a) 31 (b)

p2 16 2

y

I

I

x

33.

(a)

Answer Key 7 8 9 (a) (b) (b) 22 23 24 (d) (c) (a)

10 (c) 25 (a)

11 (a) 26 (d)

12 (d) 27 (a)

13 (c) 28 (c)

14 (b) 29 (b)

15 (b) 30 (a)

EBD_7764 P-122

1.

Physics

(b) When current is passed through a spring then current flows parallel in the adjacent turns.

The force exerted due to this magnetic field on current element i2 dl is dF = i2 dl B sin 90° \ dF = i2 dl

NOTE When two wires are placed

parallel to each other and current flows in the same direction, the wires attract each other. Similarly, here the various turns attract each other and the spring will compress.

2.

3.

4.

(a) We know that the magnetic field produced by a current carrying circular coil of radius m0 I ´ 2p r at its centre is B = 4p r m I Here B A = 0 ´ 2p 4p R BA m0 2I =1 ´ 2p Þ and BB = BB 4p 2 R (a) When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius. p r= qB where q = Charge of the particle p = Momentum of the particle B = Magnetic field Here p, q and B are constant for electron and proton, therefore the radius will be same. (c) Magnetic field due to current in wire 1 at point P distant r from the wire is m i B = 0 1 [ cos q + cos q] 4p r

5.

6.

é m 0 i1 cos q ù m0 ê 2 p r ú = 2 pr i1 i2 dl cos q ë û (a) The time period of a charged particle (m, q)

moving in a magnetic field (B) is T = 2pm qB The time period does not depend on the speed of the particle. (b) The workdone, dW = Fds cosq The angle between force and displacement is 90° . Therefore work done is zero. ×

×

×

×

×

×

F ×

×

7.

S (a) The situation is shown in the figure. FE= Force due to electric field FB = Force due to magnetic field It is given that the charged particle remains moving along X-axis (i.e. undeviated). Therefore FB = FE

Þ qvB = qE ÞB=

E 104 = = 103 weber/m2 v 10

Y

Z'

q i1

i2 r

B

P dl

E FB

q

m 0 i1 cos q (directed perpendicular to 2p r the plane of paper, inwards)

v E

B=

Z

X

Moving Charges and Magnetism 8. (b) Using Ampere’s law at a distance r from axis, B is same from symmetry. r uur i.e., B ´ 2pr = m 0i ò B.dl = m0i Here i is zero, for r < R, whereas R is the radius \ B=0 9. (b) Magentic field at the centre of a circular coil of radius R carrying current i is m i B= 0 2R Given : n ´ (2pr ') = 2pR ...(1) Þ nr ' = R n.m 0i B' = ...(2) 2r ' nm 0 i.n = n2 B From (1) and (2), B ' = 2pR 10. (c) The magnetic field at a point on the axis of a circular loop at a distance x from centre is,

B=

11.

12.

B2 =

B=

14. (b) 15. (b)

16. (a)

B.( x 2 + a2 )3/ 2 a3

54(53 ) Put x = 4 & a = 3 Þ B ' = = 250 µT 3´3´ 3 (a) Force between two long conductor carrying current, m 2I I F = 0 1 2 ´l 4p d µ 2(2 I1 ) I 2 F'=- 0 l 4p 3d F ' -2 \ = F 3 (d) (1)

m0 i2

-2

=

m0 ´ 4 ´ 10 4p

2(2p ´ 10 ) m B12 + B22 = 0 . 5 × 102 4p

Þ B = 10 - 7 ´ 5 ´ 10 2 Þ B = 5 × 10–5 Wb / m2 13. (c) Equating magnetic force to centripetal force,

m 0i a 2

2( x 2 + a 2 )3 / 2 m i B' = 0 2a

\ B' =

P-123

2

17. (d)

mv 2 = qvB sin 90º r Time to complete one revolution, 2pr 2pm = T= v qB Due to electric field, it experiences force and accelerates i.e. its velocity decreases. The charged particle will move along the lines of electric field (and magnetic field). Magnetic field will exert no force. The force by electric field will be along the lines of uniform electric field. Hence the particle will move in a straight line. B2 m0 n2i2 = B1 m 0 n1i1 i 100 ´ B2 3 Þ = 6.28 ´ 10 -2 200 ´ i 6.28 ´ 10 -2 Þ B2 = = 1.05 ´ 10 -2 Wb/m 2 6 Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the amperean æ aö path formed at a distance r1 ç = ÷ è 2ø

a/2

P1

P2

(2)

The magnetic field due to circular coil, B1 µ i m 0 i1 m ´ 3 ´ 10 = 01 = = 0 2 2r 4p 2 (2 p ´ 10 )

2

æ p r2 ö = ç 12 ÷ ´ I , where I is total current èpa ø \ Magnetic field at P1 is

EBD_7764 P-124

Physics

B1 =

m0 ´ current enclosed Path

P

A

D

B1

æ p r2 ö m 0 ´ ç 12 ÷ ´ I è pa ø m ´ I r1 = 0 Þ B1 = 2p r1 2p a 2 Now, magnetic field at point P2,

B2 d

I1

I2

O

m0 I m I . = 0 . B2 = 2p (2a) 4pa \ Required ratio =

18.

19.

a 2 r1 2 ´ 2 = = = 1. a a (d) There is no current inside the pipe. ur uur Therefore oò B . d l = µo I I=0\ B=0 r r (b) Here, E and B are perpendicular to each r other and the velocity v does not change; therefore E qE = qvB Þ v = B Also, r r E´B E B sin q = 2 B B2

E r = |v| =v B B (b) When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant). Therefore, the tangential momentum will change at every point. But kinetic energy 1 will remain constant as it is given by mv 2 2 and v2 is the square of the magnitude of velocity which does not change. (c) Clearly, the magnetic fields at a point P, equidistant from AOB and COD will have directions perpendicular to each other, as they are placed normal to each other. =

20.

21.

B1 m 0 Ir1 4pa = ´ B2 2pa 2 m 0 I

E B sin 90° 2

C

B

\ Resultant field, B = B12 + B22

But B1 =

m 0 I1 m I and B2 = 0 2 2 pd 2 pd 2

æ m ö \ B= ç 0 ÷ è 2pd ø

(I

(

2 1

m0 2 I1 + I 22 2 pd 22. (c) The magnetic field is

or, B =

B=

+ I 22

)

)

1/ 2

m0 2I 2 ´ 100 = 10 -7 ´ = 5 × 10–6 T 4p r 4

W

N 100A

=

4m E

S Ground B

According to right hand palm rule, the magnetic field is directed towards south. 23. (a) The magnetic field at O due to current in DA is mo I p ´ (dir ected vertically 4p a 6 upwards) The magnetic field at O due to current in BC is B1 =

P-125

Moving Charges and Magnetism

= q éë3i$ + $j + 2kˆ + $i ( -12 - 1)

mo I p ´ (dir ected vertically 4p b 6 downwards) The magnetic field due to current AB and CD at O is zero. Therefore the net magnetic field is B2 =

- $j ( -9 - 1) + k ( 3 - 4 ) ùû

= q éë3iˆ + $j + 2k$ - 13i$ + 10$j - k$ ùû = q éë -10i$ + 11j$ + k$ ùû Fy = 11qjˆ

B = B1 - B2 (directed vertically upwards)

=

24.

25.

26.

mo I p m o I p ´ 4 p a 6 4p b 6

m I æ 1 1ö m I = o ç - ÷ = o (b - a) 24 è a b ø 24ab r r r (d) F = I ( l ´ B) The force on AD and BC due to current I1 is zero. This is because the directions of uur r current element I d l and magnetic field B are parallel. (a) The magnetic field varies inversely with the distance for a long conductor. That is, 1 Bµ d so, graph in option (a) is the correct one. (d)

Thus, the y component of the force. 28.

(c)

a

\ Magnetic dipole moment (M) M =

dq Current in a small element, dI = I p

Magnetic field due to the element m 2dI dB = 0 4p R

Magnetic dipole moment q = 2m Angular momentum

29.

The component dB cos q, of the field is cancelled by another opposite component. Therefore,

(b)

q æ mR 2 ö 1 4 .ç ÷ .w = s.pR w. 2m è 2 ø 4

mv2 mv = qvB Þ r = r qB mp vp ; Þ rp = qpB rd =

md v d m v ; ra = a a qd B qa B

ma = 4mp , md = 2mp qa = 2qp , q d = qp

dB

Bnet = ò dB sin q =

27.

m0 I

p

ò

2p 2 R 0

sin qd q =

From the problem 1 Ep = Ed = Ea = mp vp 2 2 1 1 = md vd 2 = ma va2 2 2

m0I

p2 R

(a) Lorentz force acting on the particle ur ur r ur F = q éë E + v × B ùû ˆi ˆj kˆ ù é ê ú = q ê 3i$ + $j + 2k$ + 3 4 1 ú ê ú ê 1 1 -3 ú ë û

Þ vp2 = 2vd2 = 4mv22 Thus we have, ra = rp < rd 30.

(a) As we know, Magnetic flux, f = B. A

EBD_7764 P-126

m0 (2)(20 ´ 10 -2 ) 2 2[(0.2)2 + (0.15) 2 ]

31.

´ p(0.3 ´ 10 -2 ) 2

On solving = 9.216 × 10–11 = 9.2 × 10–11 weber (b) Work done in moving the conductor is,

33.

(c)

Physics

r r F1 = F2 = 0 because of action and reaction pair

34.

(b)

2

W = ò Fdx

Case (a) :

bA =

0

µ0 I µ I ´ 2p = 0 ´ 2p 4p R 4p l / 2p

(2pR = l)

2

= ò 3.0 ´ 10 -4 e -0.2 x ´ 10 ´ 3dx 0

=

µ0 I ´ (2p) 2 4p l

Case (b) :

l=3m I = 10 A

z

a x

= =

a/2

2 9 ´ 10 -3 e-0.2 x dx 0

ò

9 ´ 10 0.2

-3

BB = 4 ×

[-e -0.2 ´ 2 + 1]

B = 3.0 ´ 10 function)

-4 -0.2 x

e

P=

32.

W t

2.97 ´ 10-3

= 2.97 W (0.2) ´ 5 ´ 10-3 r r (a) For stable equilibrium M || B r r For unstable equilibrium M || (–B)

µ0 I [sin 45° + sin 45°] 4p a / 2

µ0 I 2 µ0 I 64 m0 I ´ ´ = ´ = 32 2 4p l / 8 2 4p l 2 4pl [4a = l] Ig G = ( I – Ig)s \ 10–3 × 100 = (10 – 10–3) × S \ S » 0.01W = 4´

(By exponential

9 ´ 10-3 ´ [1 - e -0.4 ] = 0.2 = 9 × 10–3 × (0.33) = 2.97 × 10–3J Power required to move the conductor is,

P=

45° BB

35.

(c)

36. (c) Given : Current through the galvanometer, ig = 5 × 10–3 A Galvanometer resistance, G = 15W Let resistance R to be put in series with the galvanometer to convert it into a voltmeter. V = ig (R + G) 10 = 5 × 10–3 (R + 15) \ R = 2000 – 15 = 1985 = 1.985 × 103 W

P-127

Moving Charges and Magnetism

19

Magnetism and Matter 1.

A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of T' oscillation is T ' , the ratio is T

[2003]

1 2 1 (c) 2 (d) 4 A magnetic needle lying parallel to a magnetic field requiers W units of work to turn it through

(a)

2.

1

5.

2 2

(b)

600 . The torque needed to maintain the needle in this position will be [2003]

(a)

3W

6.

(b) W

3 (d) 2 W W 2 The magnetic lines of force inside a bar magnet [2003] (a) are from north-pole to south-pole of the magnet (b) do not exist (c) depend upon the area of cross-section of the bar magnet (d) are from south-pole to north-pole of the Magnet Curie temperature is the temperature above which [2003] (a) a ferromagnetic material becomes paramagnetic (b) a paramagnetic material becomes diamagnetic (c)

3.

4.

7.

8.

(c) a ferromagnetic material becomes diamagnetic (d) a paramagnetic material becomes ferromagnetic The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2s. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be [2004] 2 s (a) 2 3 s (b) 3 2 s (c) 2 s (d) 3 The materials suitable for making electromagnets should have [2004] (a) high retentivity and low coercivity (b) low retentivity and low coercivity (c) high retentivity and high coercivity (d) low retentivity and high coercivity A magnetic needle is kept in a non-uniform magnetic field. It experiences [2005] (a) neither a force nor a torque (b) a torque but not a force (c) a force but not a torque (d) a force and a torque Needles N 1 , N 2 and N 3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will [2006] (a) attract N1 and N2 strongly but repel N3 (b) attract N1 strongly, N2 weakly and repel N3 weakly (c) attract N1 strongly, but repel N2 and N3 weakly (d) attract all three of them

EBD_7764 P-128

9.

10.

11.

1 (b)

1.

Physics

Relative permittivity and permeability of a material er and mr, respectively. Which of the following values of these quantities are allowed for a diamagnetic material? [2008] (a) er = 0.5, mr = 1.5 (b) er = 1.5, mr = 0.5 (c) er = 0.5, mr = 0.5 (d) er = 1.5, mr = 1.5 Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth.s magnetic induction is 3.6× 10.5Wb/m2) [2013] (a) 3.6 × 10.5 Wb/m2 (b) 2.56 × 10.4 Wb/m2 (c) 3.50 × 10.4 Wb/m2 (d) 5.80 × 10.4 Wb/m2 The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 103 Am–1.

2 (a)

3 (d)

4 (a)

5 (b)

6 (b)

The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is: [2014] (a) 30 mA (b) 60 mA (c) 3 A (d) 6 A 12. Hysteresis loops for two magnetic materials A and B are given below : [2016] D B

H

These materials are used to make magnets for elecric generators, transformer core and electromagnet core. Then it is proper to use : (a) A for transformers and B for electric generators. (b) B for electromagnets and transformers. (c) A for electric generators and trasformers. (d) A for electromagnets and B for electric generators

A ns we r K e y 7 8 9 (d) (b) (b)

I mBH where I = Moment of inertia of the rectangular magnet m = Magnetic moment BH = Horizontal component of the earth’s magnetic field Case 1

Case 2

10 (b)

11 (c)

12 (b)

Magnet is cut into two identical pieces such that each piece has half the original

by T = 2p

I 1 I = M l2 where mBH 12

(B)

(A)

(b) The time period of a rectangular magnet oscillating in earth’s magnetic field is given

T = 2p

H

I' m ' BH

length. Then T ' = 2p

2

where I ' =

\ 2.

(a)

T' = T

1 æ M ö æ lö I m çè ÷ø çè ÷ø = and m ' = 12 2 2 8 2 I' m ´ = m' I

I /8 m ´ = m/2 I

W = MB (cos q1 - cos q2 ) = MB (cos 0° - cos 60°)

1 1 = 4 2

Moving Charges and Magnetism

1 MB = MB(1 - ) = 2 2 MB = 3W 2 (d) As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

\ t = MB sin q = MB sin 60° = 3 3.

N

S

4.

(a) The temperature above which a ferromagnetic substance becomes paramagnetic is called Curie’s temperature.

5.

(b)

T = 2p

I I = 2p M ´B MB

where

1 I= ml 2 12 When the magnet is cut into three pieces the pole strength will remain the same and 1 æ mö æ l ö I M.I. (I¢) = çè ÷ø çè ÷ø ´ 3 = 12 3 3 9 We have, Magnetic moment (M) = Pole strength (m) × l \ New magnetic moment, æ lö M ' = m ´ ç ÷ ´ 3 = ml = M è 3ø

2 s. 9 3 (b) Electromagnet should be amenable to magnetisation & demagnetization. \ retentivity should be low & coercivity should be low. (d) A magnetic needle kept in non uniform agnetic field experience a force and torque due to unequal forces acting on poles. \T'=

6.

7.

T

=

P-129 (b) Ferromagnetic substance has magnetic domains whereas paramagnetic substances have magnetic dipoles which get attracted to a magnetic field. Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to their orbital motion of electrons these substances are repelled. 9. (b) For a diamagnetic material, the value of µr is less than one. For any material, the value of Îr is always greater than 1. Solving we get, B = 5 × 10–8 tesla 10. (b) Given : M1 = 1.20 Am2 N BH B1 S S B2 O N N r r

8.

S M2 = 1.00 Am2 ; r = Bnet = B1 + B2 + BH

Bnet = =

m 0 ( M1 + M 2 ) + BH 4p r3

10 -7 (1.2 + 1) (0.1)3

20 cm = 0.1m 2

+ 3.6 ´ 10 -5 = 2.56 ´ 10 -4 wb/m2

11. (c) Magnetic field in solenoid B = m0n i B Þ m = ni 0 (Where n = number of turns per unit length) B Ni 100i 3 = m0 L Þ 3 ´ 10 = 10 ´ 10-2 Þ i = 3A 12. (b) Graph [A] is for material used for making permanent magnets (high coercivity) Graph [B] is for making electromagnets and transformers.

Þ

EBD_7764 P-130

Physics

20

Electromagnetic Induction 1.

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is [2002]

L

2.

3.

4.

5.

v

V

(a) zero (b) RvB (c) vBL/R (d) vBL Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon [2003] (a) the rates at which currents are changing in the two coils (b) relative position and orientation of the two coils (c) the materials of the wires of the coils (d) the currents in the two coils When the current changes from +2 A to –2A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self -induction of the coil is [2003] (a) 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is 0.2×10–4T, then the e.m.f. developed between the two ends of the conductor is [2004] (a) 5 mV (b) 50 mV (c) 5 mV (d) 50mV

One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure . If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, l and v where l is the width of each tube, will be [2005] A B

6.

7.

8.

9.

v

X

C (a) – Blv (b) Blv (c) 2 Blv (d) zero The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of [2005] (a) 8 mF (b) 4 mF (c) 2 mF (d) 1 mF In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency w in a magnetic field B. The maximum value of emf generated in the coil is [2006] (a) N.A.B.R.w (b) N.A.B (c) N.A.B.R. (d) N.A.B.w The flux linked with a coil at any instant 't' is given by f = 10t2 – 50t + 250. The induced emf at t = 3s is [2006] (a) –190 V (b) –10 V (c) 10 V (d) 190 V Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is [2008] (m0 = 4p × 10 –7 Tm A–1) (a) 2.4p × 10–5 H (b) 4.8p × 10–4 H –5 (c) 4.8p × 10 H (d) 2.4p × 10–4 H

Electromagnetic Induction 10. A boat is moving due east in a region where the earth's magnetic field is 5.0 × 10–5 NA–1 m–1 due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms–1, the magnitude of the induced emf in the wire of aerial is: [2011] (a) 0.75 mV (b) 0.50 mV (c) 0.15 mV (d) 1mV 11. A horizontal straight wire 20 m long extending from east to west falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field 0.30 × 10–4 Wb/m2. The instantaneous value of the e.m.f. induced in the wire will be [2011 RS] (a) 3 mV (b) 4.5 mV (c) 1.5 mV (d) 6.0 mV 12. A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to : [2012] (a) developement of air current when the plate is placed (b) induction of electrical charge on the plate (c) shielding of magnetic lines of force as aluminium is a paramagnetic material. (d) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. 13. A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity w. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure : [2012]

1 (d)

2 (b)

3 (d)

4 (b)

5 (c)

6 (d)

P-131

(a)

B

(b) B R R

(c)

B

(d)

B

R

R 14. A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed w on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is [2013]

(a)

2 Bwl2 2

(b)

3Bwl 2 2

4 Bwl2 5Bwl 2 (d) 2 2 15. In a coil of resistance 100W, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is [2017] (c)

(a) 250 Wb (c) 200 Wb

Answer Key 7 8 9 (d) (b) (d)

10 (c)

11 (a)

(b) 275 Wb (d) 225 Wb

12 (d)

13 (a)

14 (d)

15 (a)

EBD_7764 P-132

1.

Physics

(d) The induced emf is r r -d f d ( B. A) - d ( BA cos 0º ) e= == dt dt dt ×

×

×

×

×

×

×

×

×

× l

×

×

×

×

×

×

×

×

×

8.

V

dA d (l ´ x ) = -B dt dt

10.

dx = - Blv dt (b) Mutual inductance depends on the relative position and orientation of the two coils.

11.

\ e = - Bl

2. 3.

(d)

e=-

Df -D ( LI ) DI = = -L Dt Dt Dt

4 DI Þ 8= L ´ \ | e |= L 0.05 Dt Þ L=

4.

8 ´ 0.05 = 0.1H 4

6.

Bwl 0.2 ´ 10 ´ 5 ´ 1 = = 50mV 2 2 (c) Relative velocity = v + v = 2v \ emf. = B.l (2v) (d) For maximum power, XL = XC, which yields C=

7.

13.

x

-4

e=

5.

12.

14.

(b) l = 1m, w = 5 rad/s, B = 0.2 ´ 10 -4 T

4p ´ 10-7 ´ 300 ´ 400 ´ 100 ´ 10 -4 0.2 m N N A M= 0 1 2 l = 2.4p × 10–4 H (c) Induced emf = vBH l = 1.5 × 5 × 10–5 × 2 = 15 × 10–5 = 0.15 mV (a) W E eind = Bvl = 0.3 × 10–4 × 5 × 20 = 3 × 10–3 V = 3 mV. (d) Because of the Lenz's law of conservation of energy. (a) The magnetic field due to a disc is given as m wQ 1 B= 0 i.e., B µ 2 pR R (d) Here, induced e.m.f. w l 2l =

X

\ e = –B

df = - (20t - 50) dt et = 3 = –10 V m0 N1 N2 A (d) M = l e=-

9.

×

(b) f = 10t2 – 50t + 250

1 (2 pn) 2 L

=

1 4 p 2 ´ 50 ´ 50 ´ 10

\ C = 0.1 ´ 10-5 F = 1mF ur ur df d ( N B. A) =(d) e = dt dt

d ( BA cos wt ) = NBAw sin wt dt Þ e max = NBAw = -N

e=

3l

dx

ò (wx) Bdx = Bw

2l

5 Bl 2 w = 2 15. (a) According

[(3l)2 – (2l)2 ] 2

to Faraday's

electromagnetic induction, Also, e = iR

law of df e= dt

df Þ ò d f = R ò idt dt Magnitude of change in flux (df) = R × area under current vs time graph

\

iR =

or,

1 1 df = 100 ´ ´ ´ 10 = 250 Wb 2 2

21

Alternating Current 1.

2.

The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity w is [2002]

6.

(a) R/ w L (b) R/(R2 + w 2L2)1/2 (c) w L/R (d) R/(R2 – w 2L2)1/2 The inductance between A and D is [2002] 7. A

3H

3H

3H

D

Alternating current can not be measured by D.C. ammeter because [2004] (a) Average value of current for complete cycle is zero (b) A.C. Changes direction (c) A.C. can not pass through D.C. Ammeter (d) D.C. Ammeter will get damaged. In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50V. The voltage across the LC combination will be[2004] (a) 100 V

3.

4.

5.

(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H. In a transformer, number of turns in the primary coil are 140 and that in the secondary coil are 280. If current in primary coil is 4 A, then that in the secondary coil is [2002] (a) 4 A (b) 2 A (c) 6 A (d) 10 A. In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is [2003] Q Q (a) (b) 3 2 Q (c) (d) Q 2 The core of any transformer is laminated so as to [2003] (a) reduce the energy loss due to eddy currents (b) make it light weight (c) make it robust and strong (d) increase the secondary voltage

(b)

50 2 V

(c) 50 V (d) 0 V (zero) In a LCR circuit capacitance is changed from C to 2 C. For the resonant frequency to remain unchanged, the inductance should be changed from L to [2004] (a) L/2 (b) 2 L (c) 4 L (d) L/4 9. The phase difference between the alternating p current and emf is . Which of the following 2 cannot be the constituent of the circuit? [2005] (a) R, L (b) C alone (c) L alone (d) L, C 10. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be [2005] (a) 0.4 (b) 0.8 (c) 0.125 (d) 1.25 11. A coil of inductance 300 mH and resistance 2 W is connected to a source of voltage 2V. The current reaches half of its steady state value in [2005] (a) 0.1 s (b) 0.05 s (c) 0.3 s (d) 0.15 s 8.

EBD_7764 P-134

12.

13.

Physics

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kW with C = 2mF. The resonant frequency w is 200 rad/s. At resonance the voltage across L is [2006] (a) 2.5 × 10–2 V (b) 40 V (c) 250 V (d) 4 × 10–3 V An inductor (L = 100 mH), a resistor (R = 100 W) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is [2006]

the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is [2009] 12 -3t -t / 0.2 e V V (a) (b) 6 1 - e t (c) 12e–5t V (d) 6e–5t V 17. In the circuit shown below, the key K is closed at t = 0. The current through the battery is [2010]

(

K

V

L

L

)

R1

R2

R A

B

(a)

E

14.

15.

16.

(a) 1/eA (b) eA (c) 0.1 A (d) 1 A In an a.c. circuit the voltage applied is E = E0 sin wt. The resulting current in the circuit is pö æ I = I 0 sin ç wt - ÷ . The power consumption è 2ø in the circuit is given by [2007] E I (b) P = 0 0 (a) P = 2 E0 I0 2 E0 I 0 (c) P = zero (d) P = 2 An ideal coil of 10H is connected in series with a resistance of 5W and a battery of 5V. 2second after the connection is made, the current flowing in ampere in the circuit is [2007] (a) (1 – e–1) (b) (1 – e) (c) e (d) e–1 E L R1 R2 S

An inductor of inductance L = 400 mH and resistors of resistance R1 = 2W and R2 = 2W are connected to a battery of emf 12 V as shown in

VR1R2 R12

+ R22

V at t = 0 and R at t = ¥ 2

(b)

V ( R1 + R2 ) V at t = 0 and at t = ¥ R1 R2 R2

(c)

V at t = 0 and R2

(d)

V ( R1 + R2 ) V at t = 0 and at t = ¥ R1 R2 R2

VR1 R2 R12 + R22

at t = ¥

18. In a series LCR circuit R = 200W and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is [2010] (a) 305 W (b) 210 W (c) Zero W (d) 242 W 19. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is: [2011] (a)

p LC 4 LC

(b)

2p LC

(c) (d) p LC 20. A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct

P-135

Alternating Current

supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4) [2011] (a) 21.

22.

1.7 × 105 W

(b)

A

C

L

2.7 × 106 W

(c) 3.3 × 107 W (d) 1.3 × 104 W Combination of two identical capacitors, a resistor R and a dc voltage source of voltage 6V is used in an experiment on a (C-R) circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 second. For series combination the time for needed for reducing the voltage of the fully charged series combination by half is [2011 RS] (a) 10 second (b) 5 second (c) 2.5 second (d) 20 second In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and t = RC is Capacitive time constant). Which of the following statement is correct ? [2013]

R

B

(a)

e 1- e

(b) 1

1- e e An inductor (L = 0.03 H) and a resistor (R = 0.15 kW) are connected in series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K 1 is opened and key K 2 is closed simultaneously. At t = l ms, the current in the

(c) –1

24.

(d)

circuit will be : ( e5 @ 150 )

0.03 H

[2015]

0.15 kW K2

V

S1 C

S2

L

23.

K1

15V

R

(a) Work done by the battery is half of the energy dissipated in the resistor (b) At, t = t, q = CV/2 (c) At, t = 2t, q = CV (1 – e–2) (d) At, t = 2 t, q = CV (1 – e–1) In the circuit shown here, the point ‘C’ is kept connected to point ‘A’ till the current flowing through the circuit becomes constant. Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to point ‘B’ at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to: [2014]

(a) 6.7 mA

(b) 0.67 mA

(c) 100 mA

(d) 67 mA

25. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below : [2015] L

R C

If a student plots graphs of the square of

(

)

maximum charge Q 2Max on the capacitor with time(t) for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

EBD_7764 P-136

Physics

L1

2

(a)

26. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to : [2016] (a) 0.044 H (b) 0.065 H (c) 80 H (d) 0.08 H

QMax

L2 t

27. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be : [2017]

2

(b)

QMax Q 0 (For both L1 and L2) t

L1

2

(c)

QMax

L2 t

2

(d)

QMax L 1

L2 t

1 (b) 16 (c)

1.

2 (d) 17 (c)

3 (b) 18 (d)

4 (c) 19 (a)

5 (a) 20 (b)

6 (a) 21 (c)

2

R

+w

2.

(c)

CE

XL= w L

f

3. R

R R2 + w 2 L2

r2 (r + r2 )

10 (b) 25 (c)

11 (a) 26 (b)

(b)

CE

r1 (r1 + r)

(d)

CE

r1 (r2 + r)

13 (a)

14 (c)

12 (c) 27 (a)

15 (a)

(d) These three inductors are connected in parallel. The equivalent inductance Lp is given by

1 1 1 1 1 1 1 3 = + + = + + = =1 L p L1 L2 L3 3 3 3 3

2

L

Power factor cosf =

CE

Answer Key 7 8 9 (d) (a) (a) 22 23 24 (c) (c) (b)

(b) The impedance triangle for resistance (R) and inductor (L) connected in series is shown in the figure. 2

(a)

\ Lp = 1 (b) Np = 140, Ns = 280, Ip = 4A, Is = ? Is N p For a transformer I = N p s

Þ

I s 140 = Þ Is = 2 A 4 280

Alternating Current 4. (c) When the capacitor is completely charged, the total energy in the LC circuit is with the capacitor and that energy is

1 Q2 2 C When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get E=

E 1 Q '2 = where Q ' is the charge on 2 2 C one plate of the capacitor

Q 1 1 Q 2 1 Q '2 Þ Q' = ´ = 2 2 2 C 2 C Laminated core provide less area of crosssection for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses. D.C. ammeter measure average current in AC current, average current is zero for complete cycle. Hence reading will be zero. Since the phase difference between L & C is p, \ net voltage difference across LC = 50 – 50 = 0 For resonant frequency to remain same LC should be const. LC = const \

5.

(a)

6.

(a)

7.

(d)

8.

(a)

L 2 (a) Phase difference for R–L circuit lies æ pö between ç 0, ÷ è 2ø R 12 4 = = 0.8 (b) Power factor = cos f = = Z 15 5 (a) The charging of inductance given by,

P-137

L 300 ´ 10 ´ 0.69 log 2 = R 2 Þ t = 0.1 sec. V 100 = 0.1 A 12. (c) Across resistor, I = = R 1000 At resonance,

Þ t=

1 1 = = 2500 wC 200 ´ 2 ´ 10-6 Voltage across L is I X L = 0.1 ´ 2500 = 250 V 13. (a) Initially, when steady state is achieved, X L = XC =

E R Let E is short circuited at t = 0. Then E At t = 0, i0 = R Let during decay of current at any time the current flowing is - L di - iR = 0 dt

i=

Þ

10. 11.

Rt æ - ö i = i0 ç1 - e L ÷ çè ÷ø Rt

Rt

Rt = log1 - log 2 L

i

ò

i0

t

di R = - dt i ò L 0

R

- t i R = - t Þ i = i0 e L i0 L R

14.

-100 ´10-3

1 E - t 100 100 ´10 -3 Þi= e L = = e R 100 e (c) We know that power consumed in a.c. circuit is given by, P = Erms.Irms cos f Here, E = E0 sin wt

pö æ I = I0 sin ç wt - ÷ è 2ø which implies that the phase difference, f=

i0 1 = i0 (1 - e L ) Þ e L = 2 2 Taking log on both the sides,

-

di R = - dt Þ i L

Þ log e

Þ LC = L' × 2C Þ L ' =

9.

-3

p 2

p \ P = Erms .I rms .cos =0 2 p æ ö çè Q cos = 0÷ø 2

EBD_7764 P-138

15.

16.

Physics

R æ - tö (a) I = I o ç1 - e L ÷ çè ÷ø (When current is in growth in LR circuit) 5 R æ - ´2 ö - tö 5 Eæ 10 L ç ÷ = 1 e ÷ = ç1 - e R çè ÷ø ÷ø 5 çè –1 = (1 – e ) (c) Growth in current in LR2 branch when switch is closed is given by

E [1 - e - R2t / L ] i= R2

di E R2 - R2t / L E = = e . .e dt R2 L L Hence, potential drop across

R2t L

æ E -R t / L ö -R t / L L = çè e 2 ÷ø L = Ee 2 L 2t

17.

-3

= 12e 400´10 (c) At t = 0 , no current will flow through L and R1 V \ Current through battery = R 2 At t = ¥ , RR effective resistance, Reff = 1 2 R1 + R2 V \ Current through battery = R eff = 12e–5tV

V ( R1 + R2 ) R1R2 (d) When capacitance is taken out, the circuit is LR.

=

18.

\ tan f =

wL R

1 200 = 3 3 Again, when inductor is taken out, the circuit is CR.

Þ wL = R tan f = 200 ´

\ Þ

tan f =

1 w CR

1 1 200 = R tan f = 200 ´ = wc 3 3

2

2

=

æ 200 200 ö (200) 2 + ç ÷ = 200 W 3ø è 3

Power dissipated = Vrms I rms cos f = Vrms . =

Þ

-

ö 2 æ 1 - wL ÷ Now, Z = R + ç è wC ø

Vrms R . Z Z

V 2rms R

Rö æ çQ cos f = ÷ Zø è

(220)2 ´ 200

= (200)2 Z2 220 ´ 220 = 242 W = 200

19. (a) Energy stored in magnetic field = Energy stored in electric field = \

1 2 Li 2

1 q2 2 C

1 2 1 q2 Li = 2 2 C

Also q = q0 cos wt and w =

1 LC

p LC On solving t = 4

20. (b) We have, V = V0 (1 – e–t/RC) Þ 120 = 200(1 – e–t/RC) Þ t = RC in (2.5) Þ R = 2.71 × 106 W 21. (c) Time constant for parallel combination = 2RC Time constant for series combination RC = 2 In first case :

V = V0 e

t - 1 2 RC

=

In second case : V = V0 e

-

t2 ( RC / 2)

From (1) and (2) t1 t2 = 2 RC ( RC / 2)

V0 2

=

...(1)

V0 2

t 10 Þ t2 = 1 = = 2.5 sec. 4 4

....(2)

P-139

Alternating Current 22. (c) Charge on he capacitor at any time t is given by q = CV (1– et/t) at t = 2t q = CV (1 – e–2) 23. (c) Applying Kirchhoff's law of voltage in closed loop V –VR –VC = 0 Þ R = -1 VC

A

d 2q

R dq q + =0 L dt Lc dt From damped harmonic oscillator, the 2

amplitude is given by A = Ao e Double 2

d x

VR

C

R

dt

I(0) =

15 ´ 100

0.15 ´ 10 I (¥) = 0

3

–t L e /R

–t

+ i( ¥ )

R

0.15 ´1000

I(t) = 0.1 e 0.03 = 0.67mA (c) From KVL at any time t

R

+

L

R 2 + X 2L

10 =

I(t) = 0.1 e L / R = 0.1 e L

-

Rt 2L

2

Þ Q max =

Rt 2 L Qoe

di dt –

+ – q c q di - iR - L = 0 c dt

dq q dq Ld 2q Þ + R+ 2 =0 dt c dt dt

=

e R 2 + w2 L2

=

e R 2 + 4p2 v 2 L2

220 64 + 4p2 (50) 2 L

V 80 = =8] I 10 On solving we get L = 0.065 H 27. (a) In steady state, flow of current through capacitor will be zero. Current through the circuit,

[Q R =

i=

i

i=-

equation

b dx k + = x 0 m dt m

e

i=

= 0.1A

I (t) = [I (0) – I (¥)]

25.

+

dt 2m

Hence damping will be faster for lesser self inductance. 26. (b) Here

B

(b)

2

differential

Qmax = Q oe

L VL

24.

+

E r + r2

Potential difference through capacitor Vc =

æ E ö Q = E - ir = E - ç r C è r + r2 ÷ø

\

Q = CE

r2 r + r2

EBD_7764 P-140

Physics

22

Electromagnetic Waves 1.

2.

3.

4.

5.

Electromagnetic waves are transverse in nature is evident by [2002] (a) polarization (b) interference (c) reflection (d) diffraction An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a dielectric medium with permittivity Î = 4.0. Then [2004] (a) wave length is halved and frequency remains unchanged (b) wave length is doubled and frequency becomes half (c) wave length is doubled and the frequency remains unchanged (d) wave length and frequency both remain unchanged. An electromagnetic wave in vacuum has the r r electric and magnetic field E and B , which are always perpendicular to each other. The r direction of polarization is given by X and that r of wave propagation by k . Then [2012] r r r r r (a) X || B and k || B ´ E r r r r r (b) X || E and k || E ´ B r r r r r (c) X || B and k || E ´ B r r r r r (d) X || E and k || B ´ E The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is : [2013] (a) 3 V/m (b) 6 V/m (c) 9 V/m (d) 12 V/m During the propagation of electromagnetic waves in a medium: [2014]

6.

(a) Electric energy density is double of the magnetic energy density. (b) Electric energy density is half of the magnetic energy density. (c) Electric energy density is equal to the magnetic energy density. (d) Both electric and magnetic energy densities are zero. Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists: [2014] List 1 1. Infrared waves (i) 2. 3. 4.

7.

List 2 To treat muscular strain Radio waves (ii) For broadcasting X-rays (iii) To detect fracture of bones Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere

1 2 3 4 (a) (iv) (iii) (ii) (i) (b) (i) (ii) (iv) (iii) (c) (iii) (ii) (i) (iv) (d) (i) (ii) (iii) (iv) Arrange the following electromagnetic radiations per quantum in the order of increasing energy : [2016] A : Blue light B : Yellow light C : X-ray D : Radiowave. (a) C, A, B, D (b) B, A, D, C (c) D, B, A, C (d) A, B, D, C

P-141

Electromagnetic Waves

1 (a)

1. 2.

2 (a)

3 (b)

4 (b)

5 (c)

6 (d)

Answer Key 7 (c)

(a) The phenomenon of polarisation is shown only by transverse waves. (a) Frequency remains constant during refraction

vmed =

1 µ0 Î0 ´4

=

5.

(c) E0 = CB0 and C =

1 e 0 E0 2 = m E 2 1 Bo 2 = mB Magnetic energy density = 2 m0 Thus, mE = mB

l med vmed c / 2 1 = = = l air 2 vair c

3.

4.

m0 e 0

Electric energy density =

c 2

\ wavelength is halved and frequency remains unchanged (b) Q The E.M. wave are transverse in nature i.e., r r k ´E r =H = …(i) mw r r where H = B r r m r k ´H and … (ii) = -E we r r r r k is ^ H and k is also ^ to E r r r r r or In other words X || E and k || E ´ B (b) From question, B0 = 20 nT = 20 × 10–9T (Q velocity of light in vacuum C = 3 × 108 ms–1) r r r E0 = B0 ´ C r r r | E 0 |=| B | . | C |= 20 ´ 10 -9 ´ 3 ´ 108 = 6 V/m.

1

Energy is equally divided between electric and magnetic field 6.

(d) (1) Infrared rays are used to treat muscular strain because these are heat rays. (2) Radio waves are used for broadcasting because these waves have very long wavelength ranging from few centimeters to few hundred kilometers (3) X-rays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like dones. (4) Ultraviolet rays are absorbed by ozone of the atmosphere.

7.

(c)

E, Decreases g-rays X-rays uv-rays Visible rays IR rays Radio VIBGYOR Microwaves waves

Radio wave < yellow light < blue light < Xrays (Increasing order of energy)

EBD_7764 P-142

Physics

23

Ray Optics and Optical Instruments 1.

2.

3.

4.

5.

6.

An astronomical telescope has a large aperture to [2002] (a) reduce spherical aberration (b) have high resolution (c) increase span of observation (d) have low dispersion If two mirrors are kept at 60° to each other, then the number of images formed by them is [2002] (a) 5 (b) 6 (c) 7 (d) 8 Which of the following is used in optical fibres? [2002] (a) total internal reflection (b) scattering (c) diffraction (d) refraction. Consider telecommunication through optical fibres. Which of the following statements is not true? [2003] (a) Optical fibres can be of graded refractive index (b) Optical fibres are subject to electromagnetic interference from outside (c) Optical fibres have extremely low transmission loss (d) Optical fibres may have homogeneous core with a suitable cladding. The image formed by an objective of a compound microscope is [2003] (a) virtual and diminished (b) real and diminished (c) real and enlarged (d) virtual and enlarged To get three images of a single object, one should have two plane mirrors at an angle of [2003] (a) 60º (b) 90º (c) 120º (d) 30º

7.

A light ray is incident perpendicularly to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n [2004]

45° 45°

45°

8.

9.

(a)

n>

(c)

n
2

(d) n < 2 2 A plano convex lens of refractive index 1.5 and radius of curvature 30 cm, is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object [2004] (a) 60 cm (b) 30 cm (c) 20 cm (d) 80 cm A fish looking up through the water sees the outside world contained in a circular horizon. If 4 the refractive index of water is and the fish is 3 12 cm below the surface, the radius of this circle in cm is [2005] 36 (a) (b) 36 7 7 (c) 4 5 (d) 36 5

P-143

Ray Optics and Optical Instruments

10.

11.

12.

13.

A thin glass (refractive index 1.5) lens has optical power of – 5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be [2005] (a) – 1D (b) 1 D (c) – 25 D (d) 25 D The refractive index of a glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then, [2006] (a) D1 < D2 (b) D1 = D2 (c) D1 can be less than or greater than D2 depending upon the angle of prism (d) D1 > D2 Two lenses of power –15 D and +5 D are in contact with each other. The focal length of the combination is [2007] (a) + 10 cm (b) – 20 cm (c) – 10 cm (d) + 20 cm A student measures the focal length of a convex lens by putting an object pin at a distance ‘u’ from the lens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the studen t should look like [2008] v(cm)

v(cm)

(a)

(b) O

u(cm)

O

v(cm)

v(cm)

(c)

(d) O

14.

15.

u(cm)

u(cm)

O

u(cm)

An experment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by[2008] (a) a vernier scale provided on the microscope (b) a standard laboratory scale (c) a meter scale provided on the microscope (d) a screw gauge provided on the microscope A transparent solid cylindrical rod has a 2 . It is surrounded by air.. refractive index of 3 A light ray is incident at the mid-point of one end of the rod as shown in the figure.

q

The incident angle q for which the light ray grazes along the wall of the rod is : [2009] (a)

sin -1

(

3/2

)

(b)

æ 2 ö sin -1 ç ÷ è 3ø

æ 1 ö sin -1 ç ÷ (d) sin -1 (1/ 2 ) è 3ø In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coordinates of P will be [2009]

(c)

16.

æ f fö (b) ( f, f ) çè 2 , 2 ÷ø (c) ( 4 f, 4 f ) (d) ( 2 f, 2 f ) 17. Let the x-z plane be the boundary between two transparent media. Medium 1 in z ³ 0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index of 3 . A ray of light in medium 1 given by the vector r A= 6 3iˆ + 8 3 ˆj - 10 kˆ is incident on the plane of separation. The angle of refraction in medium 2 is: [2011] (a) 45° (b) 60° (c) 75° (d) 30° 18. A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is : [2011]

(a)

(a)

1 m/s 15

(b) 10 m/s 1

(c) 15 m/s (d) m/s 10 19. A beaker contains water up to a height h1 and kerosene of height h2 above water so that the total height of (water + kerosene) is (h1 + h2).

EBD_7764 P-144

Physics

Refractive index of water is m1 and that of kerosene is m2. The apparent shift in the position of the bottom of the beaker when viewed from above is [2011 RS] (a)

æ 1 çè 1 + m

ö æ 1ö ÷ø h1 - çè1 + m ÷ø h2 1 2

21.

22.

23.

i

3ö ÷ has focal length f. When it is measured 2ø

respectively. The correct relation between the focal lengths is: [2014] (a) f1 = f2 < f (b) f1 > f and f2 becomes negative (c) f2 > f and f1 becomes negative (d) f1 and f2 both become negative 25. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is µ, a ray, incident at an angle q, on the face AB would get transmitted through the face AC of the prism provided : [2015] A q

B

C

(a)

é æ æ 1 öù q > cos -1 êµsin ç A + sin -1 ç ÷ ú è µ ø ûú è ëê

(b)

é æ æ 1 öù q < cos-1 êµsin ç A + sin -1 ç ÷ ú êë è µ ø úû è

(c)

é æ æ 1 öù q > sin -1 êµsin ç A - sin -1 ç ÷ ú è µ ø ûú è ëê

(d)

é æ æ 1 öù q < sin -1 êµsin ç A - sin -1 ç ÷ ú è µ ø ûú è ëê

d

(a)

(b) o

o

i

4 5 and , it has the focal lengths f1 and f2 3 3

When monochromatic red light is used instead of blue light in a convex lens, its focal length will [2011 RS] (a) increase (b) decrease (c) remain same (d) does not depend on colour of light An object at 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distan ce (from lens) should object shifted to be in sharp focus of film? [2012] (a) 7.2 m (b) 2.4 m (c) 3.2 m (d) 5.6 m Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 × 108 m/s, the focal length of the lens is [2013] (a) 15 cm (b) 20 cm (c) 30 cm (d) 10 cm The graph between angle of deviation (d) and angle of incidence (i) for a triangular prism is represented by [2013] d

(d) o

in two different liquids having refractive indices

ö ÷ h1 ø

æ æ 1ö 1 ö (d) ç 1 - ÷ h2 + ç1 - ÷ h1 è m1 ø è m2 ø

20.

(c)

æ çm = è

ö æ 1 ö ÷ø h1 + çè1 - m ÷ø h2 1 2

(c)

d

24. A thin convex lens made from crown glass

æ 1 (b) ç 1 è m

æ æ 1 ö 1 ç 1 + ÷ h2 - ç1 + m m è 1ø è 2

d

o

i

i

P-145

Ray Optics and Optical Instruments

26.

27.

1 (b) 16 (d)

1.

2.

3.

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears : [2016] (a) 20 times taller (b) 20 times nearer (c) 10 times taller (d) 10 times nearer In an experiment for determination of refractive index of glass of a prism by i – d, plot it was found thata ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index? [2016] (a) 1.7 (b) 1.8 (c) 1.5 (d) 1.6

2 (a) 17 (a)

3 (a) 18 (a)

4 (b) 19 (b)

5 (c) 20 (a)

6 (b) 21 (d)

28. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is : [2017] (a) real and at a distance of 40 cm from the divergent lens (b) real and at a distance of 6 cm from the convergent lens (c) real and at a distance of 40 cm from convergent lens (d) virtual and at a distance of 40 cm from convergent lens.

Answer Key 7 8 9 (b) (c) (a) 22 23 24 (c) (c) (b)

(b) The resolving power of a telescope D R.P = 1.22 l where D = diameter of the objective lens l = wavelength of light. Clearly, larger the aperture, larger is the value of D, more is the resolving power or resolution. (a) When two plane mirrors are inclined at each other at an angle q then the number of the images of a point object placed between 360° - 1, the plane mirrors is q 360° is even if q 360° -1 = 5 \Number of images formed = 60º (a) In an optical fibre, light is sent through the fibre without any loss by the phenomenon of total internal reflection as shown in the figure. n2 n1 n1>n2

4. 5.

6.

7.

8.

10 (b) 25 (c)

11 (a) 26 (b)

12 (c) 27 (c)

13 (c) 28 (c)

14 (a)

15 (c)

(b) Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference. (c) A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope. 360 360 = =4 q 90 is an even number. The number of images formed is given by 360 360 n= -1 = -1 = 4 - 1 = 3 q 90 (b) The incident angle is 45°. Incident angle > critical angle, i > ic

(b)

When q = 90° then

\ sin i > sin ic or sin 45 > sin ic 1 sin ic = n 1 1 1 \ sin 45° > or > Þn> 2 n 2 n (c) The focal length(F) of the final mirror is 1 2 1 = + F fl f m

EBD_7764 P-146

Physics

Here

1 8 Þ f m = -8 ´ f a = -8 ´ - = 5 5 m 1.6 = ´ 5 = 1D Pm = fm 8

æ 1 1 1ö = ( µ - 1) ç - ÷ fl è R1 R2 ø

1 ù 1 é1 = (1.5 - 1) ê = a 30 úû 60 ë

11.

1 1 1 1 = 2´ + = F 60 30 / 2 10 \ F = 10 cm The combination acts as a converging mirror. For the object to be of the same size of mirror, u = 2F = 20 cm \

9.

(a)

1 3 sin qc = = m 4 or tan qc =

3 16 - 9

=

3

R = 7 12

(a) For a thin prism, D = (m – 1) A Since lb < lr Þ mr < mb Þ D1 < D2 12. (c) Power of combination is given by P = P1 + P2 = (– 15 + 5) D = – 10 D.

1 1 1 metre Þ f = = f P -10

Now, P =

æ1 ö \ f = - ç ´ 100÷ cm = -10 cm. è 10 ø 13. (c) This graph suggest that when u = – f, v = + µ v (cm)

R

qc qc

12 cm

f –f

Þ R=

10.

(b)

36 7

cm

1 æ 1.5 ö æ 1 1ö =ç - 1÷ ç - ÷ è ø fa 1 è R1 R2 ø

.... (i)

æ mg öæ 1 1 1ö =ç - 1÷ ç - ÷ f m è m m ø è R1 R2 ø 1 1ö æ 1.5 ö æ 1 =ç - 1÷ ç - ÷ .... (ii) è ø 1.6 fm è R1 R2 ø

æ ö f m ç 1.5 - 1 ÷ Dividing (i) by (ii), =–8 = f a ç 1.5 ÷ - 1÷ ç è 1.6 ø 1 1 Pa = - 5 = Þ fa = fa 5

u (cm)

When the object is moved further away from the lens, v decreases but remains positive. When u is at – µ , v = f. This is how image formation takes place for different positions of the object in case of a convex lens. 14. (a) To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope

Q

15. (c)

q

a

90 – a

P

Applying Snell’s law at Q

n=

sin 90° 1 = sin(90º -a ) cos a

\ cos a =

1 n

n

P-147

Ray Optics and Optical Instruments

\ sin a = 1 - cos 2 a = 1 -

1 n2

2 sin i = 3sin r

n2 - 1 ...(1) n

=

Ð r = 45°

Applying Snell’s Law at P sin q 2 n= Þ sin q = n ´ sin a = n - 1 ; sin a from (1) 2

æ 2 ö 4 1 -1 = \ sin q = ç ÷ - 1 = è 3ø 3 3

or 16.

(d)

æ 1 ö q = sin -1 ç ÷ è 3ø Experimental curve

|v|

18. (a) From mirror formula dv v 2 æ du ö 1 1 1 =- 2ç ÷ so, + = dt v u f u è dt ø

Straight line P

2

æ f ö du dv = -ç dt è u - f ÷ø dt dv 1 Þ = m/s dt 15

Þ

(2f, 2f)

19.

45°

(b)

|u|

1 1 1 For a convex lens - = v u f

when u = -a , v = + f when u = - f , v = +a

17.

f

f u As |u| increases, v decreases for |u| > f. The graph between |v| and |u| is shown in the figure. A straight line passing through the origin and making an angle of 45°with the x-axis meets the experimental curve at P (2f, 2f ). (a) Angle of incidence is given by

(6

1 – cos i = – 2

20.

m1

Water

h1

(a) We know that m R < m B

æ1 1 1ö = (m - 1) ç - ÷ f è R1 R2 ø 1 1 Þ Þ fR > fB . > fB fR (d) The focal length of the lens 1 1 1 1 1 20 + 1 21 = = + = = f v u 12 240 240 240 and

)

3iˆ + 8 3 ˆj - 10kˆ .kˆ

Ð i = 60 ° From Snell's law,

h2

é 1 ù = h2 ê1 - ú m ë 2û Thus, total apparent shift : æ æ 1ö 1 ö = h1 ç1 - ÷ + h2 ç1 - ÷ è m1 ø è m2 ø

1+

cos (p–i) =

Kerosene

é 1ù Apparent shift due to water = h1 ê1 - ú ë m1 û Apparent shift due to kerosene

Then u = -2 f , v = 2 f Also v =

m2

20

21.

EBD_7764 P-148

Physics

f =

Þ f2 = (–) ve 25. (c) When r2 = C, ÐN2Rc = 90° Where C = critical angle

240 cm 21

æ 1ö Shift = t ç 1 - ÷ m è ø

As sin C =

1 ö 1 æ 1ç 1 ÷ = 1´ 3/2 ø 3 è 1 35 cm = 3 3 Now the object distance u. 1 3 21 1 é 3 21 ù = = u 35 240 5 êë 7 48 úû 1 1 é 48 - 49 ù = u 5 êë 7 ´ 16 úû

A

Now v' = 12 -

u = –7 ×16 × 5 = – 560 cm = – 5.6 m

22.

Velocity of light in vacuum (c) \ n = Velocity of light in medium

\ n=

3 2 R = 3cm 3mm

1 = sin r2 v

N1 q

N2

Q r1

r2

P B

C

Applying snell's law at ‘R’ µ sin r2 = 1 sin90° ...(i) Applying snell's law at ‘Q’ 1 × sin q = µ sin r1 ...(ii) But r1 = A – r2 So, sin q = µ sin (A – r 2) sin q = µ sin A cos r2 – cos A [using (i)] From (1) 2 cos r2 = 1 – sin r2 = 1–

32 + (R – 3mm)2 = R2 Þ 32 + R2 – 2R(3mm) + (3mm)2 = R2 Þ R » 15 cm

23.

24.

1 æ 3 öæ 1 ö = ç –1 ÷ç ÷ Þ f = 30 cm f è 2 øè 15 ø (c) For the prism as the angle of incidence (i) increases, the angle of deviation (d) first decreases goes to minimum value and then increases. (b) By Lens maker's formula for convex lens

öæ 2 ö 1 æ m =ç - 1÷ ç ÷ f è mL øè R ø

for, m L1 =

4 , f1 = 4 R 3

5 for m L 2 = , f 2 = -5 R 3

R

1 µ2

...(iii)

...(iv)

By eq. (iii) and (iv) sin q = µsin A 1 -

1

- cos A µ2 on further solving we can show for ray not to transmitted through face AC

é –1 æ 1 ö ù q = sin–1 ê m sin(A – sin çè µ ÷ø ú ë û So, for transmission through face AC é –1 æ 1 ö ù q > sin–1 ê m sin(A – sin çè µ ÷ø ú ë û 26. (b) A telescope magnifies by making the object appearing closer. 27. (c) We know that i + e – A = d 35° + 79° – A = 40° \ A = 74°

P-149

Ray Optics and Optical Instruments

æ A + dm ö æ 74 + d m ö sin ç sin ç ÷ ÷ 2 è 2 ø= è ø But m = 74 sinA / 2 sin 2

15 cm

d ö 5 æ = sin ç 37° + m ÷ 3 è 2 ø

m max can be

5 . That is m max is less than 3

5 = 1.67 3 But dm will be less than 40° so 5 5 sin 57° < sin 60° Þ m = 1.5 3 3 (c) As parallel beam incident on diverging lens will form image at focus. \ v = –25 cm m
v0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface is doubled, both the Kmax and V0 are also doubled. Statement – 2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. Statement 1: Davisson-Germer experiment established the wave nature of electrons. Statement 2 : If electrons have wave nature, they can interfere and show diffraction.[2012] (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is false (c) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of statement 1 (d) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1 The anode voltage of a photocell is kept fixed. The wavelength l of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows : [2013]

Physics

I (a)

O

l

I (b)

O

l

I (c)

O

l

I (d)

O

l 20. The radiation corresponding to 3 ® 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to: [2014] (a) 1.8 eV (b) 1.1 eV (c) 0.8 eV (d) 1.6 eV 21. Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from the choices given below the list: [2015] Lis t-I A . Fran ck-Hertz Exp eriment B. Pho to -electric experimen t C. Davis on -Germer experimen t

Lis t-II Particle n atu re of

(i) ligh t (ii) Dis crete en erg y levels of atom (iii) W av e natu re o f electro n (iv )

Structu re o f ato m

P-159

Dual Nature of Radiation and Matter

22.

(a) (A)-(ii); (B)-(i); (C)-(iii) (b) (A)-(iv); (B)-(iii); (C)-(ii) (c) (A)-(i); (B)-(iv); (C)-(iii) (d) (A)-(ii); (B)-(iv); (C)-(iii) Radiation of wavelength l, is incident on a photocell. The fastest emitted electron has speed 3l , the v. If the wavelength is changed to 4 speed of the fastest emitted electron will be: [2016] 1

(a)

1 (c) 16 (d)

(b)

æ 3 ö2 ç ÷ ÷ vç è4ø

1.

3.

3 (a) 18 (c)

4 (b) 19 (d)

5 (d) 20 (b)

6 (a) 21 (a)

(c)

(d)

Answer Key 8 7 9 (d) (a) (a) 22 23 (c) (c)

(c) We know that work function is the energy required and energy E = hu \

2.

1

æ 4 ö2 æ 4 ö2 (c) > v çç ÷÷ (d) < v çç ÷÷ è3ø è3ø An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays.If lmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log lmin with log V is correctly represented in : [2017]

2 (a) 17 (c)

(b)

1

æ 4 ö2 ç ÷ ÷ vç è3ø 1

23.

(a)

E Na h u Na l Cu = = E Cu h uCu l Na

1 é ù êëQ u µ l for light úû l Na E Cu 4.5 2 \ l = E = 2.3 » 1 Cu Na (a) Formation of covalent bond is best explained by molecular orbital theory. (a) For one photocathode

10 (b)

11 (b)

12 (a)

13 (b)

14 (b)

15 (a)

1 2 mv ....(i) 2 1 For another photo cathode hf1 - W =

1 2 mv ....(ii) 2 2 Subtracting (ii) from (i) we get 1 1 ( hf1 - W ) – ( hf 2 - W ) = 2 mv12 - 2 mv22 m 2 2 \ h( f1 - f 2 ) = (v1 - v2 ) 2 hf 2 - W =

EBD_7764 P-160

Physics

2h ( f - f2 ) m 1 E (b) Momentum of photon = c 2E Change in momentum = c = momentum transferred to the surface (the photon will reflect with same magnitude of momentum in opposite direction) (d) From Equation K .E = hn - f slope of graph of K.E & n is h (Plank's constant) which is same for all metals (a) For the longest wavelength to emit photo electron 2

2

Þl=

\ v1 - v 2 =

4.

5.

6.

hc hc =fÞl = l f Þl=

7.

(a)

40 ´ 1.6 ´ 10

-16

= 310 nm

2

ær ö 1 I µ 2; = ç 2÷ = I 2 è r1 ø 4 r I1

I 2 ® 4 times I1

8.

When intensity becomes 4 times, no. of photoelectrons emitted would increase by 4 times, since number of electrons emitted per second is directly proportional to intensity. (d) de-Broglie wavelength, l=

h h = p 2. m.(K.E)

\ lµ

9.

(a)

6.6 ´ 10

-34

´ 3 ´ 10

8

» 10

11.

(b) As l decreases, y increases and hence the speed of photoelectron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

12. (a) Energy of a photon of frequency n is given by E = h n . Also, E = mc2, mc2 = hn hn c

Þ p=

hn c

13. (b) The path difference between the rays APB and CQD is Dx = MQ + QN = d cos i + d cos i Dx = 2d cos i A B

D

C i

P N M i Q We know that for constructive interference the path difference is nl

d

1

\ nl = 2d cos i

K. E

Also by de-broglie concept

If K.E is doubled, wavelength becomes l

l=

2 f = 6.2 eV = 6.2 ´ 1.6 ´10 -19 J

\

V = 5 volt hc - f = eV0 l

-7

m 1.6 ´ 10 (6.2 + 5) This range lies in ultra violet range. 10. (b) The order of time is nano second. -19

Þ mc =

6.63 ´ 10-34 ´ 3 ´ 108

I

=

hc f + eV0

h = p

h 2mK.E

nh 2meV

h

=

2meV

= 2d cos i

Here n =1 : V =

h

2

8med 2 cos 2 i

P-161

Dual Nature of Radiation and Matter =

(6.6 ´ 10 -34 )2 8 ´ 9.1´ 10

-31

´1.6 ´10

14.

= 50 V (b) 2d cos i = nldB

15.

(a)

-19

´ (10

-10 2

2

) ´ cos 30

l = 400 nm, hc = 1240 eV.nm, K.E. =1.68 eV We know that,

hc hc - W = K .E Þ W = - K .E l l

1240 - 1.68 = 3.1 – 1.68 = 1.42 eV 400 (d) We know that ÞW=

16.

eV0 = K max = hn - f

17.

18. 19.

20.

where, f is the work function . Hence, as n increases (note that frequency of X-rays is greater than that of U.V. rays), both V0 and Kmax increase. So statement 1 is correct (c) By Einstein photoelectric equation, Kmax = eV0 = hv – hv0 When v is doubled, Kmax and V0 become more than double. (c) (d) As l is increased, there will be a value of l above which photoelectrons will be cease to come out so photocurrent will become zero. Hence (d) is correct answer. (b) Radius of circular path followed by electron is given by,

r=

mu = qB

1 2meV = B eB

2m V e

B2 r 2 e = 0.8V 2m For transition between 3 to 2.

Þ

V=

æ 1 1 ö 13.6 ´ 5 E = 13.6 ç - ÷ = = 1.88eV è 4 9ø 36 Work function = 1.88 eV – 0.8 eV = 1.08 eV » 1.1eV 21. (a) Frank-Hertz experiment - Discrete energy levels of atom Photoelectric effect - Particle nature of light Davison - Germer experiment - wave nature of electron. 1 2 22. (c) hn02 – hn0 = mv 2 4 1 hn 0 - hn 0 = mv ' 2 \ 3 2 4 4 n - n0 n - n0 3 \ \ v' = v 3 = n - n0 n - n0 v2 v '2

\ v' > v

4 3

23. (c) In X-ray tube, l min =

hc eV

æ hc ö In l min = In ç ÷ - InV è eø Clearly, log lmin versus log V graph slope is negative hence option (c) correctly depicts.

EBD_7764 P-162

Physics

26

Atoms 1.

2.

(a) (c) 3.

the most energy?

If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is [2002] (a) 10.2 eV (b) 0 eV (c) 3.4 eV (d) 6.8 eV Which of the following atoms has the lowest ionization potential ? [2003] 14 7 40 18

N

(b)

Ar

(d)

133 55 16 8

( )

n=4 n=3 n=2

Cs

I

O

The wavelengths involved in the spectrum of

7.

2 1D

4.

5.

6.

deuterium are slightly different from that of hydrogen spectrum, because [2003] (a) the size of the two nuclei are different (b) the nuclear forces are different in the two cases (c) the masses of the two nuclei are different (d) the attraction between the electron and the nucleus is differernt in the two cases If the binding energy of the electron in a hydrogen atom is 13.6eV, the energy required to remove the electron from the first excited state of Li + + is [2003] (a) 30.6 eV (b) 13.6 eV (c) 3.4 eV (d) 122.4 eV The manifestation of band structure in solids is due to [2004] (a) Bohr’s correspondence principle (b) Pauli’s exclusion principle (c) Heisenberg’s uncertainty principle (d) Boltzmann’s law The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with

[2005]

8.

II

III

IV

n =1

(a) IV (b) III (c) II (d) I Which of the following transitions in hydrogen atoms emit photons of highest frequency? [2007] (a) n = 1 to n = 2 (b) n = 2 to n = 6 (c) n = 6 to n = 2 (d) n = 2 to n = 1 Suppose an electron is attracted towards the k origin by a force where ‘k’ is a constant and r ‘r’ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be ‘rn’ and the kinetic energy of the electron to be ‘Tn’. Then which of the following is true? [2008] (a)

Tn µ

1 n

2

, rn µ n

2

(b) Tn independent of n, rn µ n (c)

1 Tn µ , rn µ n n

1 (d) Tn µ , rn µ n 2 n

P-163

Atoms

9.

10.

11.

12.

13.

1 (c) 16 (b)

The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from : [2009] (a) 3 ® 2 (b) 4 ® 2 (c) 5 ® 4 (d) 2 ® 1 Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is : [2011] (a) 36.3 eV (b) 108.8 eV (c) 122.4 eV (d) 12.1 eV Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be : [2012] (a) 2 (b) 3 (c) 5 (d) 6 In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n – 1). If n>>1, the frequency of radiation emitted is proportional to : [2013] 1 1 (a) (b) n n2 1 1 (c) (d) 3 n3 n 2 Hydrogen ( 1 H1 ), Deuterium ( 1 H2 ), singly ionised Helium (2He4)+, and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are l1, l2, l3 and l4 respectively then approximately which one of the following is correct? [2014] (a) 4l1 = 2l 2 = 2l3 = l 4 (b) l1 = 2l 2 = 2l3 = l 4 (c) l1 = l 2 = 4l3 = 9l 4 (d) l1 = 2l 2 = 3l3 = 4l 4

2 (b)

3 (c)

4 (a)

5 (b)

6 (b)

14. As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion : [2015] (a) kinetic energy decreases, potential energy increases but total energy remains same (b) kinetic energy and total energy decrease but potential energy increases (c) its kinetic energy increases but potential energy and total energy decrease (d) kinetic energy, potential energy and total energy decrease 15. A particle A of mass m and initial velocity v m which is at 2 rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths lA to lB after the collision is [2017] lA 2 lA 1 = (a) (b) l = 2 lB 3 B

collides with a particle B of mass

lA 1 lA = (d) l = 2 lB 3 B 16. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = l1/l2, is given by [2017] (c)

(a) (c)

Answer Key 8 7 9 (b) (d) (c)

3 4 4 r= 3 r=

10 (b)

(b) (d)

11 (d)

12 (d)

1 3 2 r= 3 r=

13 (c)

14 (c)

15 (d)

EBD_7764 P-164

1.

Physics

(c) The energy of nth orbit of hydrogen is given by En = -

13.6 n2

é ù which ê 1 - 1 ú is maximum. Here n2 êë n12 n2 2 úû is the higher energy level.

eV /atom

-13.6 = -3.4 eV 4 Therefore the energy required to remove electron from n = 2 is + 3.4 eV. (b) The ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential. (c) The wavelength of spectrum is given by

For n = 2, En =

2.

3.

7.

æ 1 1 1ö = Rz 2 ç 2 - 2 ÷ çè n l n2 ÷ø 1

æ 1 1ö hn = -13.6 ç - ÷ çè n 2 n 2 ÷ø 2 1 For transition from n = 6 to n = 2,

7

1.097 ´ 10 m 1+ M where m = mass of electron M = mass of nucleus. For different M, R is different and therefore l is different where R =

4.

(a)

En = -

13.6 n

2

-13.6 æ 1 1 ö 2 æ 13.6 ö - 2÷ = ´ç ÷ ç 2 h è6 2 ø 9 è h ø For transition from n = 2 to n = 1, n1 =

1 ö 3 æ 13.6 ö -13.6 æ 1 = ´ç ÷. h çè 22 12 ÷ø 4 è h ø \ n1 > n2 n2 =

2

Z eV/atom

8.

(b) When F = k = centripetal force, then r

9.

k mv = r r Þ mv 2 = constat Þ kinetic energy is constant Þ T is independent of n. (c) It is given that transition from the state n =4 to n = 3 in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is 5 ® 4 .

For lithium ion Z = 3 ; n = 2 ( for first excited state)

En = -

13.6 2

2

´ 32 = -30.6 eV

5.

2 (b) Pauli’s exclusion principle.

6.

(b)

é 1 1 ù E = Rhc ê 2 - 2 ú êë n1 n2 úû

E will be maximum for the transition for

é 1 1 ù Clearly, ê ú is maximum for the êë n12 n2 2 úû third transition, i.e. 2 ® 1 . I transition represents the absorption of energy. (d) We have to find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options (c) and (d). Frequency is given by

P-165

Atoms

Increasing Energy

10.

n=5 n=4 n=3 n=2 n=1

(b) Energy of excitation, æ 1

1ö

æ1 1

-

1ö ÷ = 108.8 eV 32 ø

3rd excited state

4 3

2nd excited state

2

1st excited state ground state

1

Possible number of spectral lines 1444444 424444444 3

12.

The possible number of the spectral lines is given n(n - 1) 4(4 - 1) =6 = = 2 2 (d) DE = hv n=

»

13.

DE 1 ù k(2n - 1) é 1 =kê - 2ú = 2 2 h n û n (n - 1)2 ë (n - 1)

2k n

3

1 or n µ 3 n

(c) Wave number

Þ lµ

é 1 1 1 ù = RZ 2 ê - ú 2 l êë n1 n2 úû

1

Z2 By question n = 1 and n 1 = 2 Then, l1 = l2 = 4l3 = 9l4

U = –K

ze 2 k ze 2 ; T.E = – r 2 r

k ze 2 . Here r decreases 2 r m 15. (d) From question, mA = M; mB = 2 uA = V uB = 0 K.E =

velocity of B = V2

(d) For ground state, the principal quantum no. (n) = 1. There is a 3rd excited state for principal quantum number.

Energy states Pincipal quantum no. (n)

11.

2

(c)

Let after collision velocity of A = V1 and

D E = 13.6 p2 çè h - h ÷ø eV 1 2 Þ DE = 13.6 (3)2 çè

14.

Applying law of conservation of momentum, æ mö mu = mv1 + ç ÷ v2 è2ø or, 24= 2v1 + v2

....(i)

By law of collision v2 - v1 u-0 or, u = v2 – v1 e=

[Q

....(ii)

collision is elastic, e = 1]

using eqns (i) and (ii) v1 =

4 4 and v 2 = u 3 3

de-Broglie wavelength l =

h p

m 4 ´ u l A PB 2 3 =2 \ = = 4 l B PA m´ 3

16. (b)

From energy level diagram, using DE =

For wavelength l1 DE = – E – (–2E) = \

l1 =

hc l1

hc E

æ 4E ö hc For wavelength l2 DE = – E – ç - ÷ = è 3 ø l2 hc l 1 \ l2 = \r= 1 = æ Eö l2 3 çè ÷ø 3

hc l

EBD_7764 P-166

Physics

27

Nuclei 1.

2.

3.

4.

5.

6.

At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit (i) electrons (ii) protons (iii) He2+ (iv) neutrons The emission at instant can be [2002] (a) i, ii, iii (b) i, ii, iii, iv (c) iv (d) ii, iii If N0 is the original mass of the substance of half-life period t1/2 = 5 years, then the amount of substance left after 15 years is [2002] (a) N0/8 (b) N0/16 (c) N0/2 (d) N0/4 238 nucleus originally at rest, decays When a U by emitting an alpha particle having a speed ‘u’, the recoil speed of the residual nucleus is [2003] (a) 4u (b) - 4u 234 238 4u (c) (d) - 4u 234 238 A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is [2003] (a) 0.4 ln 2 (b) 0.2 ln 2 (c) 0.1 ln 2 (d) 0.8 ln 2 A nucleus with Z= 92 emits the following in a sequence: a, b - , b - a, a, a, a, a, b - , b - , a, b + , b + , a Then Z of the resulting nucleus is [2003] (a) 76 (b) 78 (c) 82 (d) 74 Which of the following cannot be emitted by radioactive substances during their decay? [2003]

7.

(a) Protons (b) Neutrinoes (c) Helium nuclei (d) Electrons In the nuclear fusion reaction 2 3 4 1 H + 1 H ® 2 He + n

given that the repulsive potential energy between the two nuclei is ~ 7.7 ´ 10 -14 J , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s Constant k = 1.38 ´ 10 -23 J/K ] [2003] (a) 8.

9.

10 7 K

(b) 10 5 K

(d) 10 9 K (c) 10 3 K A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of 2 : 1. The ratio of their nuclear sizes will be [2004] (a) 3½ : 1 (b) 1:21/3 1/3 (c) 2 :1 (d) 1:3½ The binding energy per nucleon of deuteron

( H) and helium nucleus ( He) is 1.1 MeV 2 1

4 2

and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is [2004] (a) 23.6 MeV (b) 26.9 MeV (c) 13.9 MeV (d) 19.2 MeV 27 10. If radius of the 13 Al nucleus is estimated to be 3.6 fermi then the radius of 125 52 Te nucleus be nearly [2005] (a) 8 fermi (b) 6 fermi (c) 5 fermi (d) 4 fermi 7 11. Starting with a sample of pure 66 Cu , of it 8 decays into Zn in 15 minutes. The corresponding half life is (a) 15 minutes 1 (c) 7 minutes 2

[2005] (b) 10 minutes (d) 5 minutes

P-167

Nuclei

12.

13.

The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, it I is reduced to . The thickness of lead which will 8 I will be [2005] reduce the intensity to 2 (a) 9 mm (b) 6mm (c) 12mm (d) 18mm A nuclear transformation is denoted by X (n, a ) 7 . 3 Li

Which of the following is the nucleus of element X ? [2005] (a) 14.

15.

10 5 B

(b)

11 4 Be

C6

9 5B

(c) (d) 7 When 3Li nuclei are bombarded by protons, and the resultant nuclei are 4Be8, the emitted particles will be [2006] (a) alpha particles (b) beta particles (c) gamma photons (d) neutrons The energy spectrum of b-particles [number N(E) as a function of b-energy E] emitted from a radioactive source is [2006] (a)

N(E) E

E0

(b)

N(E)

E0

(c)

(d)

E

N(E)

E0

E

N(E)

E0

16.

12

E

If the binding energy per nucleon in 73 Li and 4 2 He nuclei are 5.60 MeV and 7.06 MeV

respectively, then in the reaction p + 73 Li ¾¾ ® 2 42 He

energy of proton must be [2006] (a) 28.24 MeV (b) 17.28 MeV (c) 1.46 MeV (d) 39.2 MeV 17. The 'rad' is the correct unit used to report the measurement of [2006] (a) the ability of a beam of gamma ray photons to produce ions in a target (b) the energy delivered by radiation to a target (c) the biological effect of radiation (d) the rate of decay of a radioactive source 18. If M O is the mass of an oxygen isotope 17 8 O ,MP

and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is [2007] (a) (MO –17MN)c2 (b) (MO – 8MP)c2 (c) (MO– 8MP –9MN)c2 (d) MOc 2 19. In gamma ray emission from a nucleus [2007] (a) only the proton number changes (b) both the neutron number and the proton number change (c) there is no change in the proton number and the neutron number (d) only the neutron number changes 20. The half-life period of a radio-active element X is same as the mean life time of another radioactive element Y. Initially they have the same number of atoms. Then [2007] (a) X and Y decay at same rate always (b) X will decay faster than Y (c) Y will decay faster than X (d) X and Y have same decay rate initially 21. This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. [2008] Statement-1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and Statement-2 : For heavy nuclei, binding energy per nucleon

EBD_7764 P-168

Physics

increases with increasing Z while for light nuclei it decreases with increasing Z. (a) Statement-1 is false, Statement-2 is true (b) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (c) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (d) Statement-1 is true, Statement-2 is false 22.

(c) E2 > E1

(d)

E1 = 2 E2

24. The speed of daughter nuclei is (a)

c

(c)

c

Dm M + Dm Dm M

(b)

c

2Dm M

(d)

c

Dm M + Dm

[2010]

25. A radioactive nucleus (initial mass number A and atomic number Z emits 3 a - particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be [2010] (a)

A- Z -8 Z -4

(b)

A- Z -4 Z -8

A - Z - 12 A- Z -4 (d) Z -4 Z -2 26. The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 – t1)

(c)

between the time t2 when The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions : [2009] (i) A + B ® C + e (ii) C ® A + B + e (iii) D + E ® F + e and (iv) F® D + E + e, where e is the energy released? In which reactions is e positive? (a) (i) and (iii) (b) (ii) and (iv) (c) (ii) and (iii) (d) (i) and (iv) DIRECTIONS: Questions number 23-24 are based on the following paragraph. A nucleus of mass M + Dm is at rest and decays into two daughter nuclei of equal mass

M each. 2

Speed of light is c. 23.

The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then [2010] (a) E2 = 2E1

(b) E1 > E2

and time t 1 when is :

2 of it had decayed 3

1 of it had decayed 3

[2011]

(a) 14 min (b) 20 min (c) 28 min (d) 7 min 27. After absorbing a slowly moving neutron of mass mN (momentum » 0) a nucleus of mass M breaks into two nuclei of masses m1 and 5m1 (6m1 = M + mN) respectively. If the de Broglie wavelength of the nucleus with mass m1 is l, the de Broglie wavelength of the nucleus will be [2011] (a) 5l (b) l / 5 (c) l (d) 25l 28. Statement - 1 : A nucleus having energy E1 decays by b– emission to daughter nucleus having energy E2, but the b– rays are emitted with a continuous energy spectrum having end point energy E1 – E2. Statement - 2 : To conserve energy and momentum in b– decay at least three particles must take part in the transformation. [2011 RS] (a) Statement-1 is correct but statement-2 is not correct.

P-169

Nuclei

29.

1 (a) 16 (b) 31 (d)

1. 2.

(b) Statement-1 and statement-2 both are correct and statement-2 is the correct explanation of statement-1. (c) Statement-1 is correct, statement-2 is correct and statement-2 is not the correct explanation of statement-1 (d) Statement-1 is incorrect, statement-2 is correct. Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron = 1.6725 × 10–27 kg, mass of proton = 1.6725 × 10–27 kg, mass of electron = 9 × 10–31 kg). [2012] (a) 0.51 MeV

(b) 7.10 MeV

(c) 6.30 MeV

(d) 5.4 MeV

2 (a) 17 (c)

3 (c) 18 (c)

4 (a) 19 (c)

5 (b) 20 (c)

(c)

Answer Key 7 8 9 (d) (b) (a) 22 23 24 (d) (c) (b)

6 (a) 21 (d)

(a) Charged particles are deflected in magnetic field. (a) After every half-life, the mass of the substance reduces to half its initial value. 5 years

N 0 ¾¾¾¾ ® =

N0 4

5years

N0 2

¾¾¾¾ ®

5 years

¾¾¾¾ ®

N0 / 4 2

Number of half lives n =

=

30. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed number of A and B nuclei will be : [2016] (a) 1 : 4 (b) 5 : 4 (c) 1 : 16 (d) 4 : 1 31. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by [2017] T (a) t = T log (1.3) (b) t = log(1.3)

N0 / 2 2

3.

t =T

log 2 log1.3

10 (b) 25 (b)

11 (d) 26 (b)

(d)

12 (c) 27 (c)

15 =3 5

We know that n 3 N æ 1ö æ 1ö N = N = 0 N= 0ç ÷ ç ÷ 0 è 2ø è 2ø 8

log1.3 log 2

13 (a) 28 (b)

14 (c) 29 (a)

15 (c) 30 (b)

(c) Here, conservation of linear momentum can be applied 4 He 2

234

Th

90

N0 8

t =T

238 × 0 = 4 u + 234 v \ v=-

4 u 234

4 r \ speed =| v |= u 234

EBD_7764 P-170

4.

(a)

Physics

A 1 5000 1 l = log e o = log e 5 1250 t A

10. (b)

2 log e 2 = 0.4 log e 2 5 (b) The number of a-particles released = 8 Therefore the atomic number should decrease by 16 The number of b–-particles released = 4 Therefore the atomic number should increase by 4. Also the number of b+ particles released is 2, which should decrease the atomic number by 2. Therefore the final atomic number = Z –16 + 4 – 2 = Z –14 = 92 – 14 = 78 (a) The radioactive substances emit a particles (Helium nucleus), b-particles (electrons) and neutrinoes. (d) The average kinetic energy per molecule

\

=

5.

6.

7.

=

11.

(d)

8.

3 -14 kT = 7.7 ´ 10 2 2 ´ 7.7 ´ 10

-14 -23

= 3.7 ´ 10

9

æ m ö æv ö v Þ ç 1 ÷ = ç 2 ÷ given 1 = 2 v2 è m2 ø è v1 ø m1 1 = Þ m2 2

3

æ r1 ö æ 1 ö 1/ 3 1 = Þ ç r ÷ = çè 2 ÷ø 3 è 2ø r2 2

r1

(a) The chemical reaction of process is 212 H ®24 He

Energy released = 4 × (7) – 4(1.1) = 23.6 MeV

1/ 3

=

3 5

5 ´ 3.6 = 6 fermi 3

7 of Cu decays in 15 minutes. 8 7 1 æ1ö = =ç ÷ 8 8 è2ø

3

\ No. of half lifes = 3 t 15 n= or 3 = T T

Þ T = half life period =

15 = 5 minutes 3

N = N0 (1 – e–lt) Þ

N0 - N N0

= e -lt

\

1 -lt =e 8

3 ´ 0.693 = 0.1386 15

Half-lifeperiod, 0.693 0.693 t½ = = = 5 minutes l 0.1386 12. (c)

m1v1 = m2 v2

9.

æ 27 ö =ç è 125 ÷ø

3 ln 2 = lt or l =

3 ´ 1.38 ´ 10 (b) From conservation of momentum

Þ

1/ 3

\ Cu undecayed = N = 1 –

This kinetic energy should be able to provide the repulsive potential energy

ÞT =

æA ö =ç 1÷ R2 è A2 ø R1

R2 =

3 kT 2

\

1/ 3

R = R0 ( A)

Intensity I = I 0 . e - md , Applying logarithm on both sides, æ I ö - md = log ç ÷ è I0 ø

æ I /8ö - m ´ 36 = logç ÷ ..........(i) è I ø æ I / 2ö ...........(ii) -m ´ d = log ç è I ÷ø Dividing (i) by (ii), æ 1ö æ 1ö log ç ÷ 3log ç ÷ 36 è 8ø è 2ø 36 = = = 3 or d = 3 d æ 1ö æ 1ö log ç ÷ log ç ÷ è 2ø è 2ø = 12 mm

P-171

Nuclei

13.

(a)

ZX

A

On comparison, A = 7 + 4 – 1 = 10, z = 3 + 2 – 0 = 5 It is boron 5B10 7 3 Li

+ 11p ¾ ¾®

8 4 Be

0 0g

14.

(c)

15.

(c) The range of energy of b-particles is from zero to some maximum value. (b) Let E be the energy of proton, then E + 7 ´ 5.6 = 2 ´ [4 ´ 7.06]

16.

17.

18.

19.

20.

+

Þ E = 56.48 - 39.2 = 17.28MeV (c) The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to 100 ergs of energy absorbed by 1 gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist L. H. Gray); one gray equals 100 rad. (c) Binding energy = [ZMP + (A – Z)MN – M]c2 = [8MP + (17 – 8)MN – M]c2 = [8MP + 9MN – M]c2 = [8MP + 9MN – Mo]c2 (c) There is no change in the proton number and the neutron number as the g-emission takes place as a result of excitation or deexcitation of nuclei. g-rays have no charge or mass. (c) According to question,

Half life of X, T1/2 = tav , average life of Y Þ

æ dN ö -ç = l y N0 è dt ÷ø y

+ 0 n1 ¾¾ ® 3 Li7 + 2 He 4

0.693 1 = lX lY

Þ l X = (0.693). l Y

\ lX < lY . Now, the rate of decay is given by æ dN ö -ç = l X N0 è dt ÷ø x

Y will decay faster than X. 21. (d) We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement (1) is correct. The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing Z and for light nuclei, B.E/nucleon increases with increasing Z. 22. (d) For A + B ® C + e, e is positive. This is because Eb for C is greater than the Eb for A and B. Again for F ® D + E + e, e is positive. This is because Eb for D and E is greater than Eb for F. 23. (c) In nuclear fission, the binding energy per nucleon of daughter nuclei is always greater than the parent nucleus. 24. (b) By conservation of energy,

2.M 2 1 2M 2 c + . v , 2 2 2 where v is the speed of the daughter nuclei M 2 2 Þ Dmc = v 2

( M + Dm) c2 =

2 Dm M (b) As a result of emission of 1 a-particle, the mass number decreases by 4 units and atomic number decreases by 2 units. And by the emission of 1 positron the atomic number decreases by 1 unit but mass number remains constant. \ Mass number of final nucleus = A – 12 Atomic number of final nucleus = Z – 8 \ Number of neutrons = (A – 12) – (Z – 8) =A–Z –4 Number of protons = Z – 8

\v=c

25.

\ Required ratio =

A-Z -4 Z -8

EBD_7764 P-172

26.

Physics

– 9 × 10–31 kg

(b) Number of undecayed atom after time t2 ; N0 = N 0 e -lt2 3

= The energy released during the process E = Dmc2 E = 9 × 10–31× 9 × 1016 = 81 × 10–15 Joules

...(i)

Number of undecayed atom after time t1; 2N0 = N 0 e -lt1 3 -lt From (i), e 2 =

...(ii) 30.

1 3

æ 1ö

Þ–lt2 = loge çè ÷ø 3 From (ii) – e -lt2 =

...(iii) 2 3

æ 2ö

Þ–lt1 = loge çè 3 ÷ø

27.

...(iv)

Solving (iii) and (iv), we get t2 – t1 = 20 min (c) pi = 0 p f = p1 + p 2 pi = p f

0 = p1 + p 2 p1 = - p 2 l1 =

h p1

l2 =

h p2

l1 = l 2 28.

29.

l1 = l 2 = l. (b) Statement-1: Energy of b-particle from 0 to maximum so E1 - E2 is the continuous energy spectrum. Statement-2: For energy conservation and momentum conservation at least three particles, daughter nucleus, b–1 and antineutron are required. (a)

1 0n

¾¾ ® 11H + -1e 0

+n+Q The mass defect during the process

Dm = mn - mH - me = 1.6725 × 10–27 – (1.6725 × 10–27+ 9 × 10–31kg)

31.

81 ´ 10 -15

= 0.511MeV 1.6 ´ 10 –19 (b) For At½ = 20 min, t = 80 min, number of half lifes n = 4 No \ Nuclei remaining = 4 . Therefore nuclei 2 decayed No = N0 - 4 2 For Bt½ = 40 min., t = 80 min, number of half lifes n = 2 No \ Nuclei remaining = 2 . Therefore 2 nuclei decayed No = N0 - 2 2 No 1 No - 4 116 2 \Required ratio = No = 1 = No - 2 14 2 15 4 5 ´ = 16 3 4 (d) Let initially there are total N0 number of nuclei

E=

At time t

NB = 0.3(given) NA

Þ NB = 0.3NA N0 = NA + NB = NA + 0.3NA N0 1.3 As we know Nt = N0e – lt

\

or,

NA =

N0 1.3

= N0e – lt

1 = e–lt 1.3

or, t =

Þ ln(1.3) = lt

ln(1.3) ln (1.3) ln(1.3) T = ln(2) Þ t= ln(2) l T

Semiconductor Electronics : Materials, Devices and Simple Circuits 1.

2.

3.

4.

5.

6.

At absolute zero, Si acts as [2002] (a) non-metal (b) metal (c) insulator (d) none of these By increasing the temperature, the specific resistance of a conductor and a semiconductor [2002] (a) increases for both (b) decreases for both (c) increases, decreases (d) decreases, increases The energy band gap is maximum in [2002] (a) metals (b) superconductors (c) insulators (d) semiconductors. The part of a transistor which is most heavily doped to produce large number of majority carriers is [2002] (a) emitter (b) base (c) collector (d) can be any of the above three. A strip of copper and another of germanium are cooled from room temperature to 80K. The resistance of [2003] (a) each of these decreases (b) copper strip increases an d that of germanium decreases (c) copper strip decreases an d that of germanium increases (d) each of these increases The difference in the variation of resistance with temeperature in a metal and a semiconductor arises essentially due to the difference in the [2003]

28

(a) crystal sturcture (b) variation of the number of charge carriers with temperature (c) type of bonding (d) variation of scattering mechanism with temperature 7. In the middle of the depletion layer of a reversebiased p-n junction, the [2003] (a) electric field is zero (b) potential is maximum (c) electric field is maximum (d) potential is zero 8. When npn transistor is used as an amplifier [2004] (a) electrons move from collector to base (b) holes move from emitter to base (c) electrons move from base to collector (d) holes move from base to emitter 9. For a transistor amplifier in common emitter configuration for load impedance of 1kW (h fe = 50 and hoe = 25) the current gain is [2004] (a) – 24.8 (b) – 15.7 (c) – 5.2 (d) – 48.78 10. A piece of copper and another of germanium are cooled from room temperature to 77K, the resistance of [2004] (a) copper increases and germanium decreases (b) each of them decreases (c) each of them increases (d) copper decreases and germanium increases 11. When p-n junction diode is forward biased then [2004]

EBD_7764 P-174

12.

13.

14.

15.

Physics

(a) both the depletion region and barrier height are reduced (b) the depletion region is widened and barrier height is reduced (c) the depletion region is reduced and barrier height is increased (d) Both the depletion region and barrier height are increased The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is [2005] (a) 2.5 eV (b) 1.1 eV (c) 0.7 eV (d) 0.5 eV In a common base amplifier, the phase difference between the input signal voltage and output voltage is [2005] p (a) p (b) 4 p (c) (d) 0 2 In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be [2005] (a) 25 Hz (b) 50 Hz (c) 70.7 Hz (d) 100 Hz If the lattice constant of this semiconductor is decreased, then which of the following is correct? [2006] conduction band width band gap valence band width

16.

Ec

current of 5.60 mA. The value of the base current amplification factor (b) will be [2006] (a) 49 (b) 50 (c) 51 (d) 48 17. A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by [2006] (a) Ionic bonding (b) Covalent bonding (c) Vander Waals bonding (d) Metallic bonding 18. If the ratio of the concentration of electrons to 7 that of holes in a semiconductor is and the 5 7 ratio of currents is , then what is the ratio of 4 their drift velocities? [2006] 5 4 (a) (b) 8 5 4 5 (d) (c) 7 4 19. The circuit has two oppositively connected ideal diodes in parallel. What is the current flowing in the circuit? [2006] 4W

D1 12V

(a) 1.71 A

D2

3W

2W

(b) 2.00 A

(c) 2.31 A (d) 1.33 A 20. In the following, which one of the diodes reverse biased? [2006]

Eg

+10 V

Ev

(a) All Ec, Eg, Ev increase (b) Ec and Ev increase, but Eg decreases (c) Ec and Ev decrease, but Eg increases (d) All Ec, Eg, Ev decrease In a common base mode of a transistor, the collector current is 5.488 mA for an emitter

R

(a) +5 V

(b)

–10 V R –5 V

P-175

Semiconductor Electronics : Materials, Devices and Simple Circuits

(c) R –10 V

+5 V R R

(d)

21.

24. In the circuit below, A and B represent two inputs and C represents the output. [2008]

If in a p-n junction diode, a square input signal of 10 V is applied as shown [2007] 5V

RL -5V Then the output signal across RL will be 10 V

+5V (a)

conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor? [2008] (a) It is an npn transistor with R as base (b) It is a pnp transistor with R as base (c) It is a pnp transistor with R as emitter (d) It is an npn transistor with R as collector

(b)

A C B The circuit represents (a) NOR gate

(b) AND gate

(c) NAND gate

(d) OR gate

25. The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform. [2009] (c) 22.

23.

(d) -5V -10 V Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate ? [2007] (a) The number of free electrons for conduction is significant only in Si and Ge but small in C. (b) The number of free conduction electrons is significant in C but small in Si and Ge. (c) The number of free conduction electrons is negligibly small in all the three. (d) The number of free electrons for conduction is significant in all the three. A working transistor with its three legs marked P, Q and R is tested using a multimeter. No

A B Input A Input B

Output is (a)

(b)

Y

EBD_7764 P-176

Physics

(c)

A X

(d)

26.

B

A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. [2009] D R

V

~

(a) OR gate

(b) NOT gate

(c) XOR gate (d) NAND gate 28. The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a: [2011 RS] (a) NOT gate (b) NOR gate (c) AND gate (d) OR gate 29. Truth table for system of four NAND gates as shown in figure is : [2012] A

The current (I) in the resistor (R) can be shown by :

Y

B

(a) I

(a)

(b) t

(c)

(c) I

A 0 0 1 1

B 0 1 0 1

Y 0 1 1 0

A 0 0 1 1

B 0 1 0 1

Y 1 1 0 0

(b)

(d)

B 0 1 0 1

Y 0 0 1 1

A 0 0 1 1

B 0 1 0 1

Y 1 0 1 1

30. The I-V characteristic of an LED is

(d) t

27.

A 0 0 1 1

The combination of gates shown below yields [2010]

(a)

I O

V

[2013]

P-177

Semiconductor Electronics : Materials, Devices and Simple Circuits

(b)

34. If p, q, r, s are inputs to a gate and x is its output, then, as per the following time graph, the gate is : [2016]

B G Y R

p

O

V

q r

I

s

(c)

x

O

V

V R Y (d) G B 31.

32.

33.

(a) OR (b) NAND (c) NOT (d) AND 35. Identify the semiconductor devices whose characteristics are given below, in the order (p), (q), (r), (s) : [2016]

O I

I

I

The forward biased diode connection is:[2014] (a)

+2V

–2V

(b)

–3V

–3V

(c)

2V

4V

(d)

–2V

V

(p) I

(q) Resistance

dark

+2V

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is : [2015] (a) 5.48 V/m

(b) 7.75 V/m

(c) 1.73 V/m

(d) 2.45 V/m

For a common emitter configuration, if a and b have their usual meanings, the incorrect relationship between a and b is : [2016] (a) a=

V

b 1 +b

1 1 (c) a = b+ 1

(b) a=

b2 1 +b2

b (d) a= 1 -b

V Illuminated

Intensity of light

(r) (s) (a) Solar cell, Light dependent resistance, Zener diode, simple diode (b) Zener diode, Solar cell, simple diode, Light dependent resistance (c) Simple diode, Zener diode, Solar cell, Light dependent resistance (d) Zener diode, Simple diode, Light dependent resistance, Solar cell 36. In a common emitter amplifier circuit using an np-n transistor, the phase difference between the input and the output voltages will be : [2017] (a) 135° (b) 180° (c) 45° (d) 90°

EBD_7764 P-178

1 (c) 16 (a) 31 (a)

1.

2.

Physics

2 (c) 17 (b) 32 (d)

3 (c) 18 (c) 33 (b)

4 (a) 19 (b) 34 (a)

5 (c) 20 (d) 35 (c)

6 (b) 21 (a) 36 (b)

Answer Key 7 8 9 (c) (d) (d) 22 23 24 (a) (b) (d)

(c) Pure silicon, at absolute zero, will contain all the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator. (c) Specific resistance is resistivity which is given by r=

m

ne 2 t where n = no. of free electrons per unit volume and t = average relaxation time For a conductor with rise in temperature n increases and t decreases. But the decrease in t is more dominant than increase in n resulting an increase in the value of r . For a semiconductor with rise in temperature, n increases and t decreases. But the increase in n is more dominant than decrease in t resulting in a decrease in the value of r. r 2 = r1 (1 + aDT )

For conductor a is positive \ r2 > r 1 for D T positive i.e., increase in temperature. For semi conductor a is negative \ r2 < r 1 for D T positive.

3.

4.

5.

6.

7.

10 (d) 25 (d)

11 (a) 26 (b)

12 (d) 27 (a)

13 (d) 28 (b)

14 (d) 29 (a)

15 (c) 30 (a)

(c) The energy band gap is maximum in insulators. Because of this the conduction band of insulators is empty. (a) Emitter sends the majority charge carrriers towards the collector. Therefore emitter is most heavily doped. (c) The resistance of metal (like Cu) decreases with decrease in temperature whereas the resistance of a semi-conductor (like Ge) increases with decrease in temperature. (b) When the temperature increases, certain bounded electrons become free which tend to promote conductivity. Simultaneously number of collisions between electrons and positive kernels increases (c) It can be seen from the following graph -

Electric field p n Distance

8. 9.

(d) Holes move from base to emmitter. (d) In common emitter configuration current gain Ai =

=

- hf e 1 + boe RL

-50

1 + 25 ´ 10-6 ´ 1 ´ 103 = – 48.78

P-179

Semiconductor Electronics : Materials, Devices and Simple Circuits

10.

11. 12.

(d) Copper is a conductor, so its resistance decreases on decreasing temperature as thermal agitation decreases,; whereas germanium is semiconductor therefore on decreasing temper ature resistance increases. (a) Both the depletion region and barrier height is reduced. (d) Band gap = energy of photon of wavelength 2480 nm. So, DE =

hc l

æ 6.63 ´10 - 34 ´ 3 ´108 ö 1 ÷´ =ç eV -9 -19 ç ÷ 2480 ´10 è ø 1.6 ´10

13.

14.

= 0.5 eV (d) Zero; In common base amplifier circuit, input and output voltage are in the same phase. (d) Input frequency, f = 50 Hz Þ T = 1 50 For full wave rectifier, T1 =

T 1 = 2 100

Þ f1 = 100 Hz. 15.

16.

(c) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cells in various directions is called its lattice parameters or constants. Increasing these lattice constants will increase or widen the band-gap (Eg), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec and Ev decreases. (a) IC = 5.488 mA, Ie = 5.6 mA a=

17.

a 5.488 = 49 ,b= 5. 6 1- a

(b) Van der Waal's bonding is attributed to the attractive forces between molecules of a liquid. The conductivity of semiconductors

(covalent bonding) and insulators (ionic bonding) increases with increase in temperature while that of metals (metallic bonding) decreases. 18. (c)

Ie Ih

=

ne eAve nh eAvh

Þ

v 5 7 7 ve Þ e = = ´ vh 4 4 5 vh

19. (b) D 2 is forward biased whereas D 1 is reversed biased. So effective resistance of the circuit R = 4 + 2 = 6W \i =

12 =2A 6

20. (d) p-side connected to low potential and nside is connected to high potential. 21. (a) The current will flow through RL when the diode is forward biased. 22. (a) Si and Ge are semiconductors but C is an insulator. Also, the conductivity of Si and Ge is more than C because the valence electrons of Si, Ge and C lie in third, fourth and second orbit respectively. 23. (b) It is a p-n-p transistor with R as base. None of the option is correct. 24. (d)

A C B

The truth table for the above logic gate is : A 1 1 0 0

B 1 0 1 0

C 1 1 1 0

This truth table follows the boolean algebra C = A + B which is for OR gate 25. (d) Here Y = ( A + B ) = A.B = A × B . Thus, it is an AND gate for which truth table is

EBD_7764 P-180

Physics

31. (a)

A B Y

26.

27.

0

0

0

For forward bias, p-side must be at higher

0 1

1 0

0 0

potential than n-side. DV = (+ )Ve

1

1

1

32. (d) Using Uav =

(b) We know that a single p-n junction diode connected to an a-c source acts as a half wave rectifier [Forward biased in one half cycle and reverse biased in the other half cycle]. (a) The final boolean expression is,

But U av = \

(b)

( A + B) = NOR gate

\

When both inputs of NAND gate are connected, it behaves as NOT gate OR + NOT = NOR.

A B 29.

(A+B)

(A+B)

(a) Y2 = A.AB Y1 = AB

B

Y = A.AB B.AB

A

Y3 = B.AB

By expanding this Boolen expression

30.

Y = A.B + B.A Thus the truth table for this expression should be (a). (a) For same value of current higher value of voltage is required for higher frequency hence (1) is correct answer.

P 4 pr

2

E 20 =

X = ( A . B ) = A + B = A + B Þ OR gate

28.

n

P

=

1 e0 E 2 2 P

4 pr 2 ´ c

1 e0 E 2 ´ c 2 2P

4 pr 2 e 0 c

=

2 ´ 0.1 ´ 9 ´ 109 1 ´ 3 ´ 108

E0 = 6 = 2.45V/m

33. (b) We know that a =

Ic Ic and b = I Ie b

Also Ie = Ib + Ic Ic I Ic b \ a= = b = I I b + Ic 1+ b 1+ c Ib Option (b) and (d) are therefore correct. 34. (a) In case of an 'OR' gate the input is zero when all inputs are zero. If any one input is ' 1', then the output is '1'. 35. (c) Graph (p) is for a simple diode. Graph (q) is showing the V Break down used for zener diode. Graph (r) is for solar cell which shows cutoff voltage and open circuit current. Graph (s) shows the variation of resistance h and hence current with intensity of light. 36. (b) In common emitter configuration for n-p-n transistor input and output signals are 180° out of phase i.e., phase difference between output and input voltage is 180°.

29

Communication Systems 1.

2.

3.

This question has Statement – 1 and Statement – 2. Of the four choices given after the statements, choose the one that best describes the two statements. [2011] Statement – 1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals. Statement – 2 : The state of ionosphere varies from hour to hour, day to day and season to season. (a) Statement–1 is true, Statement–2 is true, Statement–2 is the correct explanation of Statement–1. (b) Statement–1 is true, Statement–2 is true, Statement–2 is not the correct explanation of Statement – 1. (c) Statement – 1 is false, Statement – 2 is true. (d) Statement – 1 is true, Statement – 2 is false. Which of the following four alternatives is not correct ? We need modulation : [2011 RS] (a) to reduce the time lag between transmission and reception of the information signal (b) to reduce the size of antenna (c) to reduce the fractional band width, that is the ratio of the signal band width to the centre frequency (d) to increase the selectivity A radar has a power of 1kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 × 106m) is : [2012] (a) 80 km (b) 16 km (c) 40 km (d) 64 km

4.

A diode detector is used to detect an amplitudemodulated wave of 60% modulation by using a condenser of capacity 250 picofarad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. [2013] D Signal

(a) 10.62 MHz (c) 5.31 MHz 5.

6.

C

R

(b) 10.62 kHz (d) 5.31 kHz

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are : [2015] (a) 2005 kHz, 2000 kHz and 1995 kHz (b) 2000 kHz and 1995 kHz (c) 2 MHz only (d) 2005 kHz and 1995 kHz Choose the correct statement : (a) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. (b) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal. (c) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. (d) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. [2016]

EBD_7764 P-182

7.

2.

3.

of the following frequencies is not contained in the modulated wave ? [2017] (a) wm + wc (b) wc – wm (c) wm (d) wc

In amplitude modulation, sinusoidal carrier frequency used is denoted by wc and the signal frequency is denoted by wm. The bandwidth (Dwm) of the signal is such that Dwm < wc. Which

2 (a)

1 (b)

1.

Physics

3 (a)

4 (b)

5 (a)

6 (c)

Answer Key 7 (c)

(b) For long distance communication, sky wave signals are used. Also, the state of ionosphere varies every time. So, both statements are correct. (a) Low frequencies cannot be transmitted to long distances. Therefore, they are super imposed on a high frequency carrier signal by a process known as modulation. Speed of electro-magnetic waves will not change due to modulation. So there will be time lag between transmission and reception of the information signal. (a) Let d is the maximum distance, upto which it can detect the objects

4.

f =

A

q

h B

5.

R

R O

6.

From DAOC

Þ

1 1 Hz = 2pma RC 2 p ´ 0.6 ´ 2.5 ´ 10 -5

4 100 ´ 10 4 ´ 10 4 Hz Hz = p 1.2 25 ´ 1.2p = 10.61 KHz This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated singnal voltage for proper detection of mdoulated signal. (a) Amplitude modulated wave consists of three frequencies are wc + wm, w,wc – wm i.e. 2005 KHz, 2000KHz, 1995 KHz (c) In amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.

=

C

d

(b) Given : Resistance R = 100 kilo ohm = 100 × 103 W Capacitance C = 250 picofarad = 250 × 10–12F t = RC = 100 × 103 × 250 × 10–12 sec = 2.5 × 107 × 10–12 sec = 2.5 × 10–5 sec The higher frequency whcih can be detected with tolerable distortion is

OC 2 = AC 2 + AO2

Audio signal

(h + R)2 = d 2 + R 2

Carrier wave

d 2 = ( h + R )2 - R 2 2

2

Amplitude modulated wave

d = (h + R) - R ; d = h + 2hR 2

d = 5002 + 2 ´ 6.4 ´ 10 6 = 80 km

7.

(c) Modulated carrier wave contains frequency wc and wc ± wm

Topic-wise Solved Papers Chemistry

Some Basic Concepts of Chemistry 1.

2.

3.

4.

5.

6.

In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is [2002] (a) C2H6N2 (b) C3H4N (c) C6H8N2 (d) C9H12N3. With increase of temperature, which of these changes? [2002] (a) molality (b) weight fraction of solute (c) molarity (d) mole fraction. Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol–1) is [2002] (a) twice that in 60 g carbon (b) 6.023 ´ 1022 (c) half that in 8 g He (d) 558.5 ´ 6.023 ´ 1023 What volume of hydrogen gas, at 273 K and 1 atm. pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen? [2003] (a) 67.2 L (b) 44.8 L (c) 22.4 L (d) 89.6 L 25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was (a) 0.14 (b) 0.28 [2003] (c) 0.35 (d) 0.07 6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [2004] (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, NA = 6.02 × 1023 mol–1)

7.

1

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3), the value of 0.1 M aqueous KOH solution required is [2004] (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL 8. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is [2004] (a) urea (b) benzamide (c) acetamide (d) thiourea 9. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture ? [2005] (a) 2.70 M (b) 1.344 M (c) 1.50 M (d) 1.20 M 10. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will [2005] (a) be a function of the molecular mass of the substance (b) remain unchanged (c) increase two fold (d) decrease twice 11. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? [2006] (a) 1.25 × 10–2 (b) 2.5 × 10–2 (c) 0.02 (d) 3.125 × 10–2

EBD_7764 C-2 12. Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is [2006] –1 (a) 2.28 mol kg (b) 0.44 mol kg–1 –1 (c) 1.14 mol kg (d) 3.28 mol kg–1 13. The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H 2 SO 4 (molar mass = 98 g mol–1) by mass will be (a) 1.45 (b) 1.64 [2007] (c) 1.88 (d) 1.22 14. In the reaction, [2007] 2Al (s) + 6HCl(aq) ® 2Al3+ (aq) + 6Cl - (aq) + 3H 2 (g) (a) 11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed (b) 6 L HCl(aq) is consumed for every 3 L H2(g) produced (c) 33.6 L H2 (g) is produced regardless of temperature and pressure for every mole Al that reacts (d) 67.2 H2(g) at STP is produced for every mole Al that reacts. 15. The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO] is added to 0.3000 dm3 of water at STP is : [2011RS] (a) 5.55 ´ 10 -4 m (b) 33.3 m (c) 3.33 × 10–2 m (d) 0.555 m 16. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g. of CO2. The empirical formula of the hydrocarbon is : [2013] (a) C2H4 (b) C3H4 (c) C6H5 (d) C7H8 17. Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which

Chemistry exists as M3+ would be : (a) 7.01% (b) 4.08%[2013] (c) 6.05% (d) 5.08% 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : [JEE M 2015] (a) 42 mg (b) 54 mg (c) 18 mg (d) 36 mg At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016] (a) C4H8 (b) C4H10 (c) C3H6 (d) C3H8 The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%) ; Carbon (22.9%), Hydrogen (10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is [JEE M 2017] (a) 15 kg (b) 37.5 kg (c) 7.5 kg (d) 10 kg 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is : [JEE M 2017] (a) 1186 (b) 84.3 (c) 118.6 (d) 11.86

18.

19.

20.

21.

Answer Key 1 (c) 16 (d)

2 (c) 17 (b)

3 (a) 18 (c)

4 (a) 19 (d)

5 (d) 20 (c)

6 (b) 21 (b)

7 (a)

8 (a)

9 (b)

10 (d)

11 (d)

12 (a)

13 (d)

14 (a)

15 (a)

C-3

Some Basic Concepts of Chemistry

1.

(c)

6.

R.N.A

Percentage Simplest ratio

C

9

9 3 = 12 4

H

1

1 =1 1

4

N

3.5

3.5 1 = 14 4

1

6.02 ´ 1020

108 n= =2 54 \ molecular formula = C6 H8N2 (c) Among all the given options molarity is correct because the term molarity involve volume which increases on increasing temperature. 558.5 = 10 moles 55.85 C (no. of moles) in 60 g of C = 60/12 = 5 moles.

3.

(a) Fe (no. of moles) =

4.

(a)

2BCl3 + 3H 2 ® 2B + 6HCl 3 or BCl3 + H 2 ® B + 3HCl 2 Now, since 10.8 gm boron requires hydrogen

=

(d) 25 × N = 0.1 × 35 ; N = 0.14 Ba(OH)2 is diacid base

M = 0.07 M

6.02 ´10 20 ´1000

= 0.01M 6.02 ´10 23 ´100 [ Q M = Moles of solute present in 1L of solution]

7.

(a)

N1V1 = N 2 V2 (Note : H3PO3 is dibasic \ M = 2N) 20 ´ 0.2 = 0.1´ V (Thus. 0.1 M = 0.2 N) \ V = 40 ml

8.

(a)

H 2SO 4 is dibasic. 0.1 MH 2SO 4 = 0.2NH 2SO 4 [ QM = 2´ N ]

Meq of H 2SO 4 taken = = 100 ´ 0.2 = 20 Meq of H 2SO 4 neutralised by NaOH = 20 ´ 0.5 = 10 Meq of H 2SO 4 neutralised by NH3 = 20 – 10 = 10 % of N2 =

1.4 ´ M eq of acid neutrialised by NH3 wt. of organic compound

=

1.4 ´ 10 = 46.6 0.3

% of nitrogen in urea =

14 ´ 2 ´ 100 = 46.6 60

Similarly % of Nitrogen in Benzamide

3 22.4 ´ ´ 21.6 < 67.2L at N.T. P.. 2 10.8

hence N = M × 2 or M =

\M =

[ Mol.wt of urea =60]

3 ´ 22.4L at N.T.P 2

hence 21.6 gm boron requires hydrogen

5.

6.02 ´ 10 23

3

Empirical formula = C3H4N (C3H4N)n = 108 (12 × 3 + 4 × 1 + 14)n = 108 (54)n = 108

2.

(b) Moles of urea present in 100 ml of sol.=

N 2

14 ´ 100 = 11.5% [C6H5CONH2 = 121] 121 14 ´ 1 ´ 100 = 23.4% Acctamide = 59 [ CH3CONH2=59] 14 ´ 2 ´ 100 = 36.8% Thiourea = 76 [NH2CSNH2 = 76] Hence the compound must be urea.

=

EBD_7764 C-4

9.

Chemistry = 3.6 × 98 g = 352.8 g \ 1000 ml solution has 352.8 g of H2SO4 Given that 29 g of H2SO4 is present in = 100 g of solution \ 352.8 g of H2SO4 is present in

(b) From the molarity equation. M1V1 + M2V2 = MV Let M be the molarity of final mixture, M=

M1V1 +M V2 2

V

100 ´ 352.8 g of solution 29 = 1216 g of solution

where V = V1 + V2

=

480 ´ 1.5 + 520 ´ 1.2 = 1.344 M 480 + 520 (d) Relative atomic mass = M=

10.

Mass of one atom of the element th

1/12 part of the mass of one atom of Carbon - 12

14.

Mass of one atom of the element

´ 12 mass of one atom of the C - 12 Now if we use 1 / 6 in place of 1 /12 the formula becomes

or

15.

11.

0.25 mole of oxygen atom º

1 ´ 0.25 mole 8

of Mg3(PO4)2 = 3.125 ´ 10 -2 mole of Mg3 (PO4)2

12.

0.01/ 60 0.01 = ; 0.3 60 ´ 0.3 d = 1 g/ml

Molality =

= 5.55 ´ 10-4 m 16. (d) Q 18 gm, H2O contains = 2 gm H \ 0.72 gm H2O contains 2 ´ 0.72 gm = 0.08 gm H 18 Q 44 gm CO2 contains = 12 gm C \ 3.08 gm CO2 contains

=

(a) æ1 M ö Apply the formula d = M ç + 2 ÷ è m 1000 ø

13.

3 ´ 22.4 L of H2 at S.T.P 6 = 11.2 L of H2 at STP (a) Molality = Moles of solute / Mass of solvent in kg

=

Relative atomic mass = Mass of one atom of element ´6 Mass of one atom of carbon \ Relative atomic mass decrease twice (d) 1 Mole of Mg3(PO4)2 contains 8 mole of oxygen atoms \ 8 mole of oxygen atoms º 1 mole of Mg3(PO4)2 mole of Mg3(PO4)2

Mass 1216 = Volume 1000 = 1.216 g/ml = 1.22 g/ml (a) 2Al(s) + 6HCl(aq) ® 2Al3+(aq) + 6Cl– (aq) + 3H2(g) Q 6 moles of HCl produces = 3 moles of H2 = 3 × 22.4 L of H2 at S.T.P \ 1 mole of HCl produces

Density =

60 ö æ1 \ 1.02 = 2.05 ç + ÷ è m 1000 ø On solving we get, m = 2.288 mol/kg (d) Since molarity of solution is 3.60 M. It means 3.6 moles of H2SO4 is present in its 1 litre solution. Mass of 3.6 moles of H2SO4 = Moles × Molecular mass

=

12 ´ 3.08 = 0.84 gm C 44

0.84 0.08 : 12 1 = 0.07 : 0.08 = 7 : 8 \ Empirical formula = C7H8 17. (b) For one mole of the oxide Moles of M = 0.98 Moles of O2– = 1 Let moles of M3+ = x \ Moles of M2+ = 0.98– x On balancing charge

\C: H=

Some Basic Concepts of Chemistry

(0.98 – x) × 2 + 3x – 2 = 0 x = 0.04 0.04 ´100 < 4.08% 0.98 (c) Let the weight of acetic acid initially be w1 in 50 ml of 0.060 N solution. Let the N = (Normality = 0.06 N) 0.06 = Þ = 0.18 g = 180 mg. After an hour, the strength of acetic acid = 0.042 N so, let the weight of acetic acid be w2 N= 0.042 = Þ w2 = 0.126 g = 126 mg So amount of acetic acid adsorbed per 3g = 180 – 126 mg = 54 mg Amount of acetic acid adsorbed per g = 18 (d) CxHy(g) + O2(g) ® xCO2(g) + H2O(l) Volume of O2 used = = 75 ml \ From the reaction of combustion 1 ml CxHy requires = 15 ml = So, 4x + y = 20 x= 3 y= 8

\ % of M3+ =

18.

19.

C3 H 8

C-5 20. (c) Percentage (by mass) of elements given in the body of a healthy human adult is :-

Oxygen = 61.4%, Carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6% Q Total weight of person = 75 kg

\ Mass due to 1H is = 75 ´

10 = 7.5 kg 100

If 1H atoms are replaced by 2H atoms. Mass gain by person would be = 7.5 kg 21. (b) Given chemical eqn M2CO3 + 2HCl ® 2MCl + H2O + CO2 1gm

0.01186 mole

from the balanced chemical eqn. nM2CO3 = nCO2 1 = 0.01186 M 2 CO3 1 \ M 2 CO3 = 0.01186 Þ M = 84.3 g/mol

EBD_7764 C-6

Chemistry

Structure of Atom 1.

2.

3.

4.

In a hydrogen atom, if energy of an electron in ground state is 13.6. ev, then that in the 2nd excited state is [2002] (a) 1.51 eV (b) 3.4 eV (c) 6.04 eV (d) 13.6 eV. Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1)? (h = 6.6 ´ 10-34 Js) [2002] –34 (a) 2.1 ´ 10–34 (b) 0.5 ´ 10 (d) 0.5 ´ 10–23. (c) 2.1 ´ 10–28 The number of d-electrons retained in Fe2+ (At. no. of Fe = 26) ion is [2003] (a) 4 (b) 5 (c) 6 (d) 3 The orbital angular momentum for an electron h revolving in an orbit is given by l (l + 1) . . 2p This momentum for an s-electron will be given by [2003] (a) zero

5.

6.

h 2p 1 h (d) + . 2 2p

(b)

h 2p Which one of the following groupings represents a collection of isoelectronic species ?(At. nos. : Cs : 55, Br : 35) [2003] (a) N3–, F–, Na+ (b) Be, Al3+, Cl– (c) Ca2+, Cs+, Br (d) Na+, Ca2+, Mg2+ In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003] (a) 5 ® 2 (b) 4 ® 1

(c)

2.

(c)

2®5

(d) 3 ® 2

7.

2

The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately [2003] (a) 10–31 metres (b) 10–16 metres (c) 10–25 metres (d) 10–33 metres Planck’s constant, h = 6.63 × 10–34 Js 8. Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004] (a) n = 4, l = 3, m = + 1, s = + ½ (b) n = 4, l = 4, m = – 4, s = – ½ (c) n = 4, l = 3, m = + 4, s = + ½ (d) n = 3, l = 2, m = – 2, s = + ½ 9. Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively [2004] (a) 16 and 4 (b) 12 and 5 (c) 12 and 4 (d) 16 and 5 10. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004] (a) 406 nm (b) 192 nm (c) 91 nm (d) 9.1×10–8 nm 11. Which one of the following sets of ions represents the collection of isoelectronic species? [2004] (a) K+, Cl–, Mg2+, Sc3+ (b) Na+, Ca2+, Sc3+, F– (c) K+, Ca2+, Sc3+, Cl– (d) Na+, Mg2+, Al3+, Cl– (Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21) 12. In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields? [2005]

Structure of Atom

13.

(A) n = 1, l = 0, m = 0 (B) n = 2, l = 0, m = 0 (C) n = 2, l = 1, m = 1 (D) n = 3, l = 2, m = 1 (E) n = 3, l = 2, m = 0 (a) (D) and (E) (b) (C) and (D) (c) (B) and (C) (d) (A) and (B) Of the following sets which one does NOT contain isoelectronic species? [2005] (a)

BO 33 - , CO 32 - , NO 3-

(b) SO32 - , CO32 - , NO3-

14.

15.

16.

17.

18.

(c)

CN - , N 2 , C 22 -

(d)

PO 34 - , SO 24 - , ClO -4

According to Bohr's theory, the angular momentum of an electron in 5th orbit is [2006] (a) 10 h / p (b) 2.5 h / p (c) 25 h / p (d) 1.0 h / p Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1, accurate upto 0.001% will be [2006] (a) 1.92 × 10–2 m (b) 3.84 × 10–2 m (c) 19.2 × 10–2 m (d) 5.76 × 10–2 m (h = 6.63 × 10–34 Js) Which one of the following sets of ions represents a collection of isoelectronic species? [2006] (a) N3–, O2–, F–, S2– (b) Li+, Na+, Mg2+, Ca2+ (c) K+, Cl–, Ca2+, Sc3+ (d) Ba2+, Sr2+, K+, Ca2+ Which of the following sets of quantum numbers represents the highest energy of an atom? [2007] (a) n = 3, l = 0, m = 0, s = +1/2 (b) n = 3, l = 1, m = 1, s = +1/2 (c) n = 3, l = 2, m = 1, s = +1/2 (d) n = 4, l = 0, m = 0, s = +1/2. Which one of the following constitutes a group of the isoelectronic species? [2008] 2– – (a) C2 , O2 , CO, NO – (b) NO+ ,C2– 2 , CN , N 2

(c)

2– CN – , N 2 ,O2– 2 , C2

(d) N 2 ,O2– , NO+ ,CO

C-7 19. The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is [2008] (a) 8.51 × 105 J mol–1 (b) 6.56 × 105 J mol–1 (c) 7.56 × 105 J mol–1 (d) 9.84 × 105 J mol–1 20. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms –1. (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 Js) [2009] (a) 0.40 nm (b) 2.5 nm (c) 14.0 nm (d) 0.32 nm 21. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ( h = 6.6 × 10–34 kg m2s–1, mass of electron, em = 9.1 × 10–31 kg) : [2009] (a) 5.10 × 10 –3 m (b) 1.92 × 10 –3 m (c) 3.84 × 10 –3 m (d) 1.52 × 10 –4 m 22. The energy required to break one mole of Cl – Cl bonds in Cl 2 is 242 kJ mol –1 . The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1). [2010] (a) 594 nm (b) 640 nm (c) 700 nm (d) 494 nm 23. Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is [2010] (a) 4.41 × 10–16 J atom–1 (b) –4.41 × 10–17 J atom–1 (c) –2.2 × 10–15 J atom–1 (d) 8.82 × 10–17 J atom–1 24. The frequency of light emitted for the transition n = 4 to n = 2 of the He+ is equal to the transition in H atom corresponding to which of the following? [2011RS] (a) n = 2 to n = 1 (b) n = 3 to n = 2 (c) n = 4 to n = 3 (d) n = 3 to n = 1 25. The electrons identified by quantum numbers n and l : [2012] (A) n = 4, l = 1 (B) n = 4, l = 0 (C) n = 3, l = 2 (D) n = 3, l = 1 can be placed in order of increasing energy as :

EBD_7764 C-8

26.

27.

(a) (C) < (D) < (B) < (A) (b) (D) < (B) < (C) < (A) (c) (B) < (D) < (A) < (C) (d) (A) < (C) < (B) < (D) The increasing order of the ionic radii of the given isoelectronic species is : [2012] – 2+ + 2– 2– – (a) Cl , Ca , K , S (b)S , Cl , Ca2+ , K+ 2+ + – 2– (c) Ca , K , Cl , S (d)K+, S2–, Ca2+, Cl– Energy of an electron is given by E = – 2.178 × æ Z2 ö 10-18 J ç 2 ÷ . Wavelength of light required to èn ø

28.

excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : [2013] (h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1) (a) 1.214 × 10–7 m (b) 2.816 × 10.–7 m (c) 6.500 × 10–7 m (d) 8.500 × 10–7 m The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is: [2014] (a)

5, 0, 0, +

(c)

5,1,1, +

1 2

1 2

(b) 5,1, 0, +

1 2

(d) 5, 0,1, +

1 2

Chemistry 29. Which of the following is the energy of a possible excited state of hydrogen? [JEE M 2015] (a) –3.4 eV (b) +6.8 eV (c) +13.6 eV (d) –6.8 eV 30. A stream of electrons from a heated filaments was passed two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/l (where l is wavelength associated with electron wave) is given by: [JEE M 2016]

(a)

(b) meV 2meV (c) meV (d) 2meV 31. The radius of the second Bohr orbit for hydrogen atom is : [JEE M 2017] (Plank's const. h = 6.6262 × 10–34 Js ; mass of electron = 9.1091 × 10–31 kg ; charge of electron e = 1.60210 × 10–19 C ; permittivity of vaccum Î0 = 8.854185 × 10–12 kg–1 m–3 A2) (a) 1.65Å (b) 4.76Å (c) 0.529Å (d) 2.12Å 32. The group having isoelectronic species is : (a) O2– , F–, Na+, Mg2+ [JEE M 2017] (b) O– , F–, Na, Mg+ (c) O2– , F–, Na , Mg2+ (d) O– , F–, Na+ , Mg2+

Answer Key 1 (a) 16 (c) 31 (d)

1.

2 (c) 17 (c) 32 (a)

4 (a) 19 (d)

5 (a) 20 (a)

6 (a) 21 (b)

7 (d) 22 (d)

8 (a) 23 (b)

9 (b) 24 (a)

10 (c) 25 (b)

11 (c) 26 (c)

(a) 2nd excited state will be the 3rd energy level.

En = 2.

3 (c) 18 (b)

13.6 n

2

eV or E =

\ Dv =

13.6 eV = 1.51 eV. 9

(c)

3. Dx. Dp =

h ; 4p

or

Dx.m.Dv =

h 4p

12 (a) 27 (a)

13 (b) 28 (a)

14 (b) 29 (a)

15 (a) 30 (b)

6.62 ´ 10 -34 4 ´ 3.14 ´ 0.025 ´ 10 -5

= 2.1 ´ 10 -28 ms -1 (c) Fe++ (26 – 2 = 24) = 1s2 2s2 2p6 3s2 3p6 4s0 3d6 hence no. of d electrons retained is 6. [Two 4s electron are removed]

C-9

Structure of Atom

4.

5. 6.

13. (b) Calculating number of electrons

(a) For s-electron, l = 0 \ Orbital angular momentum = h 0(0 + 1) = 0 2p (a) N3–, F– and Na+ contain 10 electrons each. (a) The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n1 =2, n 2 = 5 i.e. 5 ® 2. l=

(d)

8.

(a) The possible quantum numbers for 4f electron are n = 4, l = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and 1 s=± 2 Of various possiblities only option (a) is possible. (b) Electronic configuration of Cr atom (z = 24) = 1s 2 , 2s2 2p 6 ,3s 2 3p 6 3d 5 , 4s1 when l = 1, p - subshell, Numbers of electrons = 12 when l = 2, d - subshell, Numbers of electrons = 5

10.

(c)

12.

+,

20 Ca

2+ ,

-

® 15 + 8 ´ 4 + 3 = 50 ü PO34- ¾¾

4.

ï ï iso-electronic ý species ï ® 17 + 8 ´ 4 + 1 = 50ï ClO4 ¾¾ þ

2® 16 + 8 + 2 = 50 SO 4 ¾¾

Hence the species in option (b) are not iso-electronic. 14. (b) Angular momentum of an electron in nth orbital is given by ,

=

l = 91.15 ´ 10 -9 m » 91nm 19 K

3.

¾¾ ® 6 + 7 + 1 = 14ü ï ï iso-electronic ® 7 ´ 2 = 14 N 2 ¾¾ ý species ï ® 6 ´ 2 + 2 = 14 ï C 2 ¾¾ þ CN

nh 2p

For n = 5, we have Angular momentum of electron

1 æ1 1 ö = 1.097 ´ 10 7 ç - ÷ = 1.097 ´ 10 7 l è1 ¥ ø

(c)

2.

¾¾ ® 16 + 8 ´ 3 + 2 = 42ü ï ï not iso-electronic 2® 32 CO3 ¾¾ ý species ï ® 32 NO3 ¾¾ ïþ

2SO3

mvr =

æ 1 1 1 ö÷ = Rç ç n2 n2 ÷ l 2ø è 1

11.

þ

h 6.6 ´ 10 -34 = = 10 -33 m mv 60 ´ 10 -3 ´10

7.

9.

1.

3BO3 ¾¾ ® 5 + 8 ´ 3 + 3 = 32 ü ï iso-electronic 2® 6 + 8 ´ 3 + 2 = 32ý species CO3 ¾¾ ® 7 + 8 ´ 3 + 1 = 32 ï NO3 ¾¾

21Sc

3+ ,

5h 2.5h = p 2p

15. (a) Given m = 9.1 × 10–31kg h = 6.6 × 10–34Js Dv =

17 Cl

-

each contains 18 electrons. (a) The energy of an orbital is given by (n + l) in (d) and (c). (n + l) value is (3 + 2) = 5 hence they will have same energy, since there n values are also same.

300 ´ .001 100

= 0.003ms–1

From Heisenberg's uncertainity principle

Dx =

6.62 ´ 10 -34 4 ´ 3.14 ´ 0.003 ´ 9.1 ´ 10-31

= 1.92 ´ 10 -2 m

EBD_7764 C-10 16. (c) (a) N3– = 7 + 3 = 10e–, O–– ––® 8 + 2 = 10e–

F– = 9 + 1 = 10e–, S – – ––® 16 + 2 = 18e– (not iso electronic) (b) Li+ = 3+1= 4e–, Na+ = 11–1 = 10e–, Mg++ = 12–2=10e– Ca++ = 20 – 2 = 18e– (not isoelectronic) (c) K+ = 19 – 1= 18e–, Cl– =17 + 1 = 18e–, Ca++ = 20 – 2 =18e, Sc3+ = 21–3 = 18e– (isoelectronic) ++ (d) Ba 56 – 2 = 54e, Sr++ 38–2 = 36e– K+= 9–1 = 18e–, Ca++= 20–2 = 18e– (not isoelectronic) 17. (c) (a) n = 3, l = 0 means 3s-orbital and n + l= 3 (b) n = 3, l = 1 means 3p-orbital n + l = 4

(c) n = 3, l = 2 means 3d-orbital n + l = 5

18.

19.

(d) n = 4, l = 0 means 4s-orbital n + l = 4 Increasing order of energy among these orbitals is 3s < 3p < 4s < 3d \ 3d has highest energy.. (b) Species having same number of electrons are isoelectronic calculating the number of electrons in each species given here, we get. CN– (6 + 7 + 1 = 14); N2 (7 + 7 = 14); O22–(8 + 8 +2 = 18) ; C22– (6 + 6 + 2 = 14); O2– (8 + 8 + 1 = 17) ; NO+ (7 + 8 – 1 = 14) CO (6 + 8 = 14) ; NO (7 + 8 = 15) From the above calculation we find that all the species listed in choice (b) have 14 electrons each so it is the correct answer. (d) (DE), The energy required to excite an electron in an atom of hydrogen from n = 1 to n = 2 is DE (difference in energy E2 and E1) Values of E2 and E1 are, E2 =

-1.312 ´106 ´ (1)2

Chemistry DE is given by the relation, E1 = – 1.312 × 106 J mol–1 \ DE = E2 – E1 = [–3.28 × 105]– [–1.312 × 106 ] J mol–1 = (–3.28 × 105 + 1.312 × 106) J mol–1 = 9.84 × 105 J mol–1 Thus the correct answer is (d)

h 6.63 ´ 10-34 = mv 1.67 ´ 10 -27 ´ 1 ´ 103 = 3.97 × 10–10 meter = 0.397 nanometer (b) According to Heisenberg uncertainty principle.

20. (a) 21.

l=

Dx.mDv =

h 4p

Here Dv =

600 ´ 0.005 = 0.03 100

h 4 pmD v

6.6 ´ 10-34

So, Dx =

4 ´ 3.14 ´ 9.1 ´ 10-31 ´ 0.03 = 1.92 × 10–3 meter 22. (d) Energy required to break one mole of Cl – Cl bonds in Cl2

=

=

\l=

242 ´ 103 6.023 ´ 10

23

=

hc l

6.626 ´ 10-34 ´ 3 ´ 108 l 6.626 ´ 10 -34 ´ 3 ´ 108 ´ 6.023 ´ 10 23 242 ´ 108 = 494.7 nm

= 0.4947 × 10–6 m Z2

´ 13.6 eV

...(i)

I1 Z12 n22 or I = 2 ´ 2 n1 Z 2 2

...(ii)

23. (b) I. E =

n2

Given I1 = – 19.6 × 10–18 , Z1 = 2, n1 = 1 , Z2 = 3 and n2 = 1 Substituting these values in equation (ii).

(2)2

= –3.28 × 105 J mol–1

Dx =

–

19.6 ´ 10 -18 4 1 = ´ I2 1 9

C-11

Structure of Atom -18 or I2 = -19.6 ´ 10 ´

24. (a)

9 4

= – 4.41 × 10–17 J/atom For He+ 1 1ö æ 1 = RH Z 2 ç è 2 2 42 ÷ø l For H v=

v=

æ ö 1 1 1 = RH ç - ÷ ç n12 n2 ÷ l 2ø è

æ 1 1 ö hc DE = 2.178 ´ 10-18 ç 2 - 2 ÷ = è1 2 ø l 3 hc Þ 2.178 ´ 10-18 ´ = 4 l 6.62 ´ 10-34 ´ 3 ´ 108 = l 6.62 ´ 10 -34 3 ´ 108 ´ 4 l= ´3 2.178 ´ 10 -18 –7 = 1.214 × 10 m 28. (a) The electronic configuration of Rubidium (Rb = 37) is

27. (a)

For same frequency,

\

æ ö æ ö 1 1 1 1 z2 = ç 2 - 2 ÷ = ç 2 - 2 ÷ çè 2 4 ÷ø çè n1 n2 ÷ø Since, z = 2 1 1 1 1 = n12 n22 12 2 2

\ n1 = 1 & n2 = 2 25. (b) (a) 4 p (b) 4 s (c) 3 d (d) 3 p Accroding to Bohr Bury's (n + l) rule, increasing order of energy (D) < (B) < (C) < (A). Note : If the two orbitals have same value of (n + l) then the orbital with lower value of n will be filled first. 26. (c) Among isoelectronic species ionic radii increases as the charge increases. Order of ionic radii Ca2+ < K+ < Cl– < S2– The number of electrons remains the same but nuclear charge increases with increase in the atomic number causing decrease in size.

29. (a)

30. (b)

31. (d)

32. (a)

1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p 6 5s1 Since last electron enters in 5s orbital 1 Hence n = 5, l = 0, m = 0, s = ± 2 Total energy = where n = 2, 3, 4 .... Putting n = 2 ET = As electron of charge ‘e’ is passed through ‘V’ volt, kinetic energy of electron will be eV Wavelength of electron wave (l) = l = Þ\ = Radius of nth Bohr orbit in H-atom = 0.53 n2Å Radius of II Bohr orbit = 0.53 × (2)2 = 2.12 Å Isoelectronic species have same no. of electrons. ions O–2 F– Na+ Mg2+ 8+2 9+1 11–1 12–2 No. of e– = 10 10 10 10 therefore O 2– , F – , Na + , Mg 2+ are isoelectronic

EBD_7764 C-12

Chemistry

Classification of Elements and Periodicity in Properties 1.

2.

3.

4.

5.

6.

According to the Periodic Law of elements, the variation in properties of elements is related to their [2003] (a) nuclear masses (b) atomic numbers (c) nuclear neutron-proton number ratios (d) atomic masses Which one of the following is an amphoteric oxide ? [2003] (a) Na2O (b) SO2 (c) B2O3 (d) ZnO Which one of the following ions has the highest value of ionic radius ? [2004] (a) O2– (b) B3+ (c) Li+ (d) F– Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is [2004] (a) Al2O3 < SiO2< SO2 < P2O3 (b) SiO2< SO2 < Al2O3 < P2O3 (c) SO2< P2O3 < SiO2 < Al2O3 (d) Al2O3 < SiO2< P2O3 < SO2 The formation of the oxide ion O (2g-) requires first an exothermic and then an endothermic step as shown below [2004] O(g) + e - = O(g) DHº = -142 kJmol -1 2O - (g) + e - = O (g) DHº = 844 kJmol -1 This is because (a) O– ion will tend to resist the addition of another electron (b) Oxygen has high electron affinity (c) Oxygen is more elecronegative (d) O– ion has comparatively larger size than oxygen atom Which of the following oxides is amphoteric in character? [2005] (a) SnO2 (b) SiO 2 (c) CO2 (d) CaO

7.

8.

9.

3

In which of the following arrangements, the order is NOT according to the property indicated against it? [2005] (a) Li < Na < K < Rb : Increasing metallic radius (b) I < Br < F < Cl : Increasing electron gain enthalpy (with negative sign) (c) B < C < N < O Increasing first ionization enthalpy (d) Al 3+ < Mg 2+ < Na + < F Increasing ionic size Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture? [2006] (a) Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens (b) In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group (c) The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group (d) In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group In which of the following arrangements, the sequence is not strictly according to the property written against it? [2008] (a) HF < HCl < HBr , HI : increasing acid strength (b) NH3 < PH3 < AsH3 Mg 2+ > Na + > F- > O2(c) Ba < Ca < Se < S < Ar (b) Na + > Mg 2 + > Al3+ > O 2- > F(d) Ca < Ba < S < Se < Ar (c) Na + > F- > Mg2 + > O2 - > Al3+ 13. The ionic radii (in Å) of N3–, O2– and F– are (d) O2 - > F- > Na + > Mg 2 + > Al3+ respectively : [JEE M 2015] 11. The correct order of electron gain enthalpy with (a) 1.71, 1.40 and 1.36 negative sign of F, Cl, Br and I, having atomic (b) 1.71, 1.36 and 1.40 number 9, 17, 35 and 53 respectively, is : (c) 1.36, 1.40 and 1.71 [2011RS] (d) 1.36, 1.71 and 1.40 (a) F > Cl > Br > I (b) Cl > F > Br > I 14. Which of the following atoms has the highest first ionization energy? [JEE M 2016] (c) Br > Cl > I > F (d) I > Br > Cl > F (a) K (b) Sc 12. Which of the following represents the correct (c) Rb (d) Na order of increasing first ionization enthalpy for

Answer Key 1 (b)

1.

2. 3.

4.

2 (d)

3 (a)

4 (d)

5 (a)

6 (a)

7 (c)

(b) According to modern periodic law, the properties of the elements are repeated after certain regular intervals when these elements are arranged in order of their increasing atomic numbers. (d) Na2O (basic), SO2 and B2O3 (acidic) and ZnO is amphoteric. (a) O– – and F– are isoelectronic. Hence have same number of shells, therefore greater the nuclear charge smaller will be the size i.e. O– – > F– further Li + and B3+ are isoelectronic. therefore Li+ > B3+ Hence the correct order of atomic size is. O-- > F– > Li+ > B3+ (d) As the size increases the basic nature of oxides changes to acidic nature i.e., acidic nature increases. SO 2 > P 2 O 3 > SiO 2 > Al 2 O 3 Acidic

Weak acidic

Amphoteric

8 (b)

5.

6.

7.

9 (b)

10 (d)

11 (b)

12 (c)

13 (a)

14 (b)

SO 2 and P 2 O 3 are acidic as their corresponding acids H2SO3 and H3PO3 are strong acids. (a) O– ion exerts a force of repulsion on the incoming electron. The energy is required to overcome it. (a) CaO is basic as it form strong base Ca(OH)2 on reaction with water. CaO + H2O –––––® Ca(OH)2 CO2 is acidic as it dissolve in water forming unstable carbonic acid. H2O + CO2 –––––® H2CO3 Silica (SiO2) is insoluble in water and acts as a very weak acid. SnO2 is amphoteric as it reacts with both acid and base. SnO2 + 2H2SO4–––––® Sn(SO4)2 + 2H2O SnO2 + 2KOH–––––® K2SnO3 + H2O (c) In a period the value of ionisation potential increases from left to right with breaks where the atoms have some what stable configuration. In this case N has half filled

EBD_7764 C-14

8.

stable orbitals. Hence has highest ionisation energy. Thus the correct order is B< C< O < N and not as given in option (c) (b) The alkali metals are highly reactive because their first ionisation potential is very low and hence they have great tendency to loses electron to form unipositive ion. NOTE On moving down group- I from

9.

10.

Li to Cs ionisation enthalpy decreases hence the reactivity increases. The halogens are most reactive elements due to their low bond dissociation energy, high electron affinity and high enthalpy of hydration of halide ion. However their reactivity decreases with increase in atomic number (b) In hydrides of 15th group elements, basic character decreases on descending the group i.e. NH3 > PH3 > AsH3 > SbH3. (d) All the given species contains 10 e– each i.e. isoelectronic. For isoelectronic species anion having high negative charge is largest in size and the cation having high positive charge is smallest.

11.

Chemistry (b) As we move down in a group electron gain enthalpy becomes less negative because the size of the atom increases and the distance of added electron from the nucleus increases. Negative electron gain enthalpy of F is less than Cl. This is due to the fact that when an electron is added to F, the added electron goes to the smaller n = 2 energy level and experiences significant repulsion from the other electrons present in this level. In Cl, the electron goes to the larger n = 3 energy level and consequently occupies a larger region of space leading to much less electron–electron repulsion. So the correct order is

Cl > F > Br > I. 12. (c) On moving down a group size increases hence ionisation enthalpy decreases, hence Se < S and Ba < Ca. Further, Ar being an inert gas has maximum IE. 13. (a) For isoelectronic species, size of anion increases as negative charge increases. Thus the correct order is 14. (b) Alkali metals have the lowest ionization energy in each period on the other hand Sc is a d - block element. Transition metals have smaller atomic radii and higher nuclear charge leading to high ionisation energy.

C-15

Chemical Bonding and Molecular Structure

4

Chemical Bonding and Molecular Structure 1.

2.

3.

4.

5.

6.

In which of the following species the interatomic bond angle is 109° 28’? [2002] –1 + (a) NH3, (BF4) (b) (NH4) , BF3 (c) NH3, BF4 (d) (NH2)–1, BF3. Which of the following are arranged in an increasing order of their bond strengths? [2002] (a) O2– < O2 < O2+ 7 (c) a buffer solution with pH < 7 (d) a buffer solution with pH > 7. Species acting as both Bronsted acid and base is [2002] –1 (a) (HSO4) (b) Na2CO3 (c) NH3 (d) OH–1. Let the solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is [2002] 3 5 (a) 4x (b) 108x (c) 27x4 (d) 9x. Change in volume of the system does not alter which of the following equilibria? [2002] (a) N2 (g) + O2 (g) 2NO (g) (b) PCl5 (g) PCl3 (g) + Cl2 (g) (c) N2 (g) + 3H2 (g) 2NH3 (g) (d) SO2Cl2 (g) SO2 (g) + Cl2 (g). For the reactionCO (g) + (1/2) O2 (g) = CO2 (g), Kp / Kc is [2002] –1 (a) RT (b) (RT) (c) (RT)–1/2 (d) (RT)1/2 Which one of the following statements is not true? [2003] (a) pH + pOH = 14 for all aqueous solutions (b) The pH of 1 × 10–8 M HCl is 8 (c) 96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of copper at the cathode (d) The conjugate base of H 2 PO -4 is HPO 24The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product number will be [2003] (a) 4 × 10–10 (b) 1 × 10–15 (c) 1 × 10–10 (d) 4 × 10–15

8.

For the reaction equilibrium [2003] N2O4 (g) 2 NO2 (g) the concentrations of N 2 O 4 and NO 2 at equilibrium are 4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for the reaction is (a) 3 × 10–1 mol L–1 (b) 3 × 10–3 mol L–1 (c) 3 × 103 mol L–1 (d) 3.3 × 102 mol L–1 9. Consider the reaction equilibrium [2003] 2 SO2 (g) + O2(g) 2 SO3 (g) ; DHº = –198 kJ On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is (a) increasing temperature as well as pressure (b) lowering the temperature and increasing the pressure (c) any value of temperature and pressure (d) lowering of temperature as well as pressure 10. When rain is accompanied by a thunderstorm, the collected rain water will have a pH value [2003] (a) slightly higher than that when the thunderstorm is not there (b) unin fluenced by occurrence of thunderstorm (c) which depends on the amount of dust in air (d) slightly lower than that of rain water without thunderstorm. 11.

The conjugate base of H 2 PO -4 is (a) H3PO4 (c)

PO 34-

[2004]

(b) P2O5 (d) HPO 24 -

12. What is the equilibrium expression for the reaction P4 (s) + 5O2 (g) (a)

Kc = [O 2 ]5

P4 O10 (s) ?

[2004]

C-33 18. The exothermic formation of CIF3 is represented by the equation :

Equilibrium

13.

(b)

Kc = [P4O10 ] / 5[P4 ][O2 ]

(c)

Kc = [P4 O10 ]/[P4 ][O 2 ]5

(d)

Kc = 1/[O 2 ]5

Cl2 (g) + 3F2 (g)

For the reaction, COCl2 (g) the

CO(g) + Cl 2 (g)

Kp

is equal to (a) (c)

14.

Kc

[2004] (b) RT

RT 1

(d) 1.0 RT The equilibrium constant for the reaction N 2 (g) + O 2 (g)

2NO2 (g) at temperature

T is 4×10–4. The value of Kc for the reaction 1 1 N 2 (g) + O2 (g) at the same 2 2 temperature is [2004] –4 (a) 4 × 10 (b) 50 (c) 2.5 × 102 (d) 0.02 The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in term of Ksp by the relation : [2004] NO2 (g)

15.

(a) (c) 16.

17.

s = (256 K sp )1/ 5 (b) s = (128 K sp )1/ 4 1/ 4

s = ( K sp /128)

1/ 5

(d) s = ( K sp / 256)

If a is the degree of dissociation of Na 2SO 4 , the Vant Hoff’s factor (i) used for calculating the molecular mass is (a) 1 – 2 a (b) 1 + 2 a [2005] (c) 1 – a (d) 1 + a The solubility product of a salt having general formula MX 2 , in water is : 4 × 10 -12 . The concentration of M 2+ ions in the aqueous solution of the salt is [2005] (a)

4.0 ´ 10 -10 M

(c) 1.0 ´ 10

-4

M

(b) 1.6 ´ 10 -4 M (d) 2.0 ´ 10

-6

M

2ClF3 (g) ;

Δ H = – 329 kJ Which of the following will increase the quantity of CIF3 in an equilibrium mixture of Cl 2 , F2 and ClF3 ? [2005] (a) Adding F2 (b) Increasing the volume of the container (c) Removing Cl2 (d) Increasing the temperature 19. For the reaction : [2005] 2NO2(g)

2NO (g ) + O 2(g ) ,

(K c = 1.8 ´ 10- 6 at 184°C) (R = 0.0831 kJ/ (mol. K))

When K p and K c are compared at 184°C, it is found that (a) Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure (b) Kp = Kc (c) Kp is less than Kc (d) Kp is greater than Kc 20. Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be : [2005] (a)

3.98 ´ 10 -6

(b) 3.68 ´ 10 -6

(c)

3.88 ´ 106

(d) 3.98 ´ 108

21. What is the conjugate base of OH - ? [2005] (b) O (d) O 2 H2O 22. An amount of solid NH 4 HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH 3 and H 2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The (a) (c)

O2-

equilibrium constant for NH 4 HS decomposition at this temperature is [2005]

EBD_7764 C-34

23.

(a) 0.11 (b) 0.17 (c) 0.18 (d) 0.30 Phosphorus pentachloride dissociates as follows, in a closed reaction vessel [2006] PCl5(g)

PCl3(g) + Cl2(g)

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be (a)

æ x ö P (b) ç è 1 - x ÷ø

æ x ö çè ÷P x + 1ø

æ 2x ö P (d) ç è 1 - x ÷ø The equilibrium constant for the reaction

(c) 24.

æ x ö çè ÷P x - 1ø

1 SO 2 (g ) + O 2 (g) 2 is Kc = 4.9 × 10–2. The value of Kc for the reaction SO3(g)

2SO2(g) + O2(g)

25.

2SO3(g)

will be [2006] (a) 9.8 × 10–2 (b) 4.9 × 10–2 (c) 416 (c) 2.40 × 10–3 Given the data at 25ºC

Ag + I - ¾ ¾® AgI + e - E º = 0.152 V

Ag ¾ ¾® Ag+ + e -

26.

27.

28.

Eº = -0.800 V

What is the value of log Ksp for AgI? (2.303 RT/ F = 0.059 V) [2006] (a) –37.83 (b) –16.13 (c) –8.12 (d) +8.612 The first and second dissociation constants of an acid H2A are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall dissociation constant of the acid will be [2007] (a) 0.2 × 105 (b) 5.0 × 10–5 (c) 5.0 × 1015 (d) 5.0 × 10–15. The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffer solution of HA in which 50% of the acid is ionized is [2007] (a) 7.0 (b) 4.5 (c) 2.5 (d) 9.5 In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molecular mass = 283) ˆˆ† the equilibrium which sets in is AgIO3 (s) ‡ˆˆ Ag + (aq) + IO3-(aq) . If the solubility product con-

Chemistry stant Ksp of AgIO3 at a given temperature is 1.0 × 10–8, what is the mass of AgIO3 contained in 100 ml of its saturated saolution? [2007] (a) 1.0 × 10– 4 g (b) 28.3 × 10–2 g (c) 2.83 × 10–3 g (d) 1.0 × 10–7 g. 29. The equilibrium constants K p and K p2 for the 1 reactions X ƒ2Y and Z ƒ P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibria is [2008] (a) 1: 36 (b) 1 : 1 (c) 1 : 3 (d) 1 : 9 30. For the following three reactions a, b and c, equilibrium constants are given: [2008]

(i) CO(g) ∗ H 2O(g) ƒ CO 2 (g) ∗ H 2 (g); K1 (ii) CH4 (g) ∗ H2O(g) ƒ CO(g) ∗ 3H2 (g); K 2 (iii) CH4 (g) + 2H2O(g) ƒ CO2 (g) + 4H2 (g);K3 (a)

K1 K 2 = K3

(b) K 2 K3 = K1

(d) K3 .K 23 = K12 31. Four species are listed below: [2008] – + i. ii. H3 O HCO3 (c) K3 = K1 K2

iv. HSO3F iii. HSO4– Which one of the following is the correct sequence of their acid strength? (a) iv < ii < iii < i (b) ii < iii < i < iv (c) i < iii < ii < iv (d) iii < i < iv < ii 32. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the correspondng salt, BA, will be [2008] (a) 9.58 (b) 4.79 (c) 7.01 (d) 9.22 33. Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10– 4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form? (KSP for for BaCO3 = 5.1 × 10–9) [2009] (a) 5.1 × 10 –5 M (b) 8.1 × 10 –8 M (c) 8.1 × 10 –7 M (d) 4.1 × 10 –5 M 34. Three reactions involving H2PO4– are given below : [2010] (i) H3PO4 + H2O® H3O+ + H2PO4–

C-35

Equilibrium

35.

36.

37.

38.

39.

(ii) H2PO4– + H2O ® HPO42– + H3O+ (iii) H2PO4– + OH– ® H3PO4 + O2– In which of the above does H 2 PO-4 act as an acid ? (a) (ii) only (b) (i) and (ii) (c) (iii) only (d) (i) only In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11. Select the correct statement for a saturated 0.034 M solution of the carbonic acid. [2010] 2 (a) The concentration of CO3 is 0.034 M. (b) The concentration of CO32 - is greater than that of HCO3- . (c) The concentrations of H+ and HCO3- are approximately equal. (d) The concentration of H+ is double that of CO32 - . Solubility product of silver bromide is 5.0 × 10– 13. The quantity of potassium bromide (molar mass taken as 120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [2010] (a) 1.2 × 10–10 g (b) 1.2 × 10–9 g (c) 6.2 × 10–5 g (d) 5.0 × 10–8 g At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions? [2010] (a) 9 (b) 10 (c) 11 (d) 8 An acid HA ionises as ˆˆ† H + + A -1 HA ‡ˆˆ The pH of 1.0 M solution is 5. Its dissociation constant would be : [2011RS] (a) 5 (b) 5 ´ 10 -8 (d) 1 ´ 10-10 (c) 1 ´ 10-5 The Ksp for Cr(OH)3 is 1.6 × 10–30. The solubility of this compound in water is : [2011RS] (a)

40.

4 1.6 ´ 10 -30

(b) 4 1.6 ´ 10 -30 / 27 (c) 1.6 ´ 10-30/ 27 (d) 2 1.6 ´ 10 -30 The equilibrium constant (Kc) for the reaction N2(g) + O2(g) ® 2NO(g) at temperature T is 4 × 10–4. The value of Kc for the reaction [2012]

1 1 NO( g) ® N2 ( g ) + O 2 ( g ) at the same 2 2 temperature is: (a) 0.02 (b) 2.5 × 102 (c) 4 × 10–4 (d) 50.0 41. The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of the acid is : [2012] (a) 3 × 10–1 (b) 1 × 10–3 (c) 1 × 10–5 (d) 1 × 10–7 42. How many litres of water must be added to 1 litre an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ? [2013] (a) 0.1 L (b) 0.9 L (c) 2.0 L (d) 9.0 L † ˆˆˆˆˆˆ 43. For the reaction SO 2(g ) + O 2(g ) ‡ˆˆˆˆˆSO 3(g ) , if 2 x K P = KC ( RT ) where the symbols have usual meaning then the value of x is (assuming ideality): [2014] 1 (a) –1 (b) 2 1 (c) (d) 1 2 44. The standard Gibbs energy change at 300 K for the reaction 2A B + C is 2494.2 J. At a given time, the composition of the reaction mixture is [A] =, [B] = 2 and [C] = . The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718] [JEE M 2015] (a) forward direction because Q < Kc (b) reverse direction because Q < Kc (c) forward direction because Q > Kc (d) reverse direction because Q > Kc 45. The equilibrium constant at 298 K for a reaction A +B C+D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be : [JEE M 2016] (a) 1.818 (b) 1.182 (c) 0.182 (d) 0.818 46. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is [JEE M 2017] (a) 7.2 (b) 6.9 (c) 7.0 (d) 1.0 1

EBD_7764 C-36

Chemistry

Answer Key 1 (a) 16 (b) 31 (c) 46 (b)

1.

2.

3.

4.

5.

2 (a) 17 (c) 32 (c)

(a)

3 4 (a) (a) 18 19 (a) (d) 33 34 (a) (a) (i)

NOTE

5 (c) 20 (a) 35 (c)

6 (b) 21 (a) 36 (b)

7 (d) 22 (a) 37 (b)

8 (b) 23 (c) 38 (d)

K p = K c (RT)Dn ; æ 1ö 3 1 Dn = 1 - ç 1 + ÷ = 1 - = - . è 2ø 2 2

Kp

6.

10 (d) 25 (b) 40 (d)

7.

(d)

12 (d) 27 (d) 42 (d)

13 (c) 28 (c) 43 (b)

14 (b) 29 (a) 44 (d)

15 (d) 30 (c) 45 (a)

ˆˆ† A∗2 ∗ 2B, AB2 ‡ˆˆ

[A] = 1.0 × 10–5, [B] = [2.0 × 10–5] , Ksp = [B]2 [A] = [2 × 10–5]2 [1.0 × 10–5] = 4 × 10–15 8.

(b)

Kc =

[NO2 ]2 [1.2 ´ 10 -2 ]2 = [N 2O 4 ] [4.8 ´ 10 -2 ]

= 3 ´ 10–3 mol/L 9. (b) Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles. \ High pressure will be required. 10. (d) The rain water after thunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm. 11.

(d)

NOTE Conjugate acid-base differ by H+ +

–H

H 2 PO 4

-1/ 2

\ K = (RT) c (b) pH of an acidic solution should be less than 7. The reason is that from H2O. [H+] = 10– 7M which cannot be neglected in comparison to 10 –8M. The pH can be calculated as. from acid, [H+] = 10–8M. from H2O, [H+] = 10–7M \ Total [H+] = 10–8 + 10–7 = 10–8 (1 + 10) = 11× 10–8

11 (d) 26 (d) 41 (c)

\ pH = – log [H+] = –log 11×10–8 = –[log11 + 8 log 10] = –[1.0414 – 8] = 6.9586

A buffer is a solution of weak

acid and its salt with strong base and vice versa. HCl is strong acid and NaCl is its salt with strong base. pH is less than 7 due to HCl. (a) (HSO4)– can accept and donate a proton (HSO4)– + H+ ® H2SO4 (acting as base) (HSO4)– – H+ ® SO42–. (acting as acid) (a) Mg(OH)2 ® [Mg2+] + 2[OH–] x 2x Ksp = [Mg] [OH]2 = [x][2x]2 = x.4x2 = 4x3. (a) In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles. (c)

9 (b) 24 (c) 39 (b)

Acid

HPO 4 conjugate ugate base

12. (d) For P4 (s) + 5O 2 (g)

Kc =

1 (O 2 ) 5

.

The

concentration unity 13. (c)

K p = K c (RT) Dn ; Here D n = 1– 2 = –1 Kp 1 \ = K c RT

P4O10 (s) solids

have

C-37

Equilibrium

14.

(b)

Kc =

[ NO]2 = 4 ´10 - 4 [ N 2 ][O 2 ]

K ¢c =

[NO] 1

=

15.

[N 2 ]1/ 2 [O 2 ]1/ 2

4 ´ 10 -4

(d) MX4 Ksp

(b)

1 Kc

of OH– is O2– 22. (a)

M 4+ + 4X S

Vant. Hoff’s factor i =

start At equib.

0 .5 atm 0 .5 + x atm

+ H 2 S(g ) 0 atm x atm.

Þ x = 0.17 atm. p NH3 = 0.5 + 0.17 = 0.67 atm +

-–

2Na + SO 4 2a

; p H 2S = 0.17 atm

a

2 K= pNH 3 ´ pH 2S = 0.67 ´ 0.17 atm

1 - a + 2a + a = 1 + 2a 1

(c)

ˆˆ † 2NO 2 (g) ‡ˆ ˆ 2NO(g) + O 2 (g) Given Kc = 1.8 × 10–6 at 184 ºC R = 0.0831 kj/mol. k Kp= 1.8 × 10–6 × 0.0831 × 457 = 6.836 × 10– 6

[ Q 184°C = (273 + 184) = 457 k, Dn = (2 + 1, –1) = 1] Hence it is clear that Kp > Kc 20.

NH 3(g )

Then 0.5 + x + x = 2x + 0.5 = 0.84 (given)

MX 2 M + + + 2X s 2s Where s is the solubility of MX2 then Ksp = 4s3; s × (2s)2 = 4×10–12 = 4s3; s = 1 × 10–4 \ [M++] = s = 1[M++] = 1.0 × 10– 4 18. (a) The reaction given is an exothermic reaction thus accordingly to Lechatalier’s principle lowering of temperature, addition of F2 and or Cl2 favour the for ward direction and hence the production of ClF3 . 19. (d) For the reaction:17.

NH 4 HS(s )

4S

1/ 5

1- a

[H + ]

OH - ¾ ¾® H + + O 2 - Conjugate base

= 50

Na 2SO 4

1

On solving, [H+] = 3.98 × 10–6 21. (a) Conjugate acid-base pair differ by only one proton.

= [s] [4s]4 = 256 s5

æ Ksp ö \s = ç ÷ è 256 ø

16.

=

5.4 = log

(a) pH = –log [H+] = log

= 0.1139 = 0.11 23. (c)

PCl5(g)

x

x

Total moles after dissociation 1–x+x +x =1+x p = mole fraction of PCl 3

æ x ö PCl3 × Total pressure = ç ÷P è1+ x ø 1 SO 2 (g ) + O 2 (g) 24. (c) SO3(g) 2

Kc =

[SO 2 ] [O2 ]½ = 4.9 ´ 10-2 ; [SO3 ]

On taking the square of the above reaction [SO 2 ]2 [O 2 ] [SO 3 ]2

= 24.01´10 - 4

now K'C for 2SO2(g) + O2(g)

1 [H + ]

PCl3(g) + Cl2(g)

1–x

=

[SO 3 ]2 2

[SO 2 ] [O 2 ]

=

1 24.01´10 - 4

2SO3 = 416

EBD_7764 C-38

25.

Chemistry

(b) (i)

+

Ag ¾ ¾® Ag + e

-

E º = - 0.800 V

¾® AgI + e - E º = 0 .152 V (ii) Ag + I - ¾

\ s = K sp =

= 1.0 × 10–4 mol/lit = 1.0 × 10–4 × 283 g/lit

(Q Molecular mass of Ag IO3 = 283)

From (i) and (ii) we have, AgI ¾ ¾® Ag + + I - E º = -0.952 V

Eocell =

\ –0.952 =

0.059 n 0.059

1 [ Q k = [Ag+][I–]]

= log K

log [Ag+] [I–]

0.952 = log K sp or –16.13 = log or 0.059

Ksp 26. (d)

ˆˆ† H∗ ∗ HA, H 2 A ‡ˆˆ + \ K = 1.0 × 10–5 = [H ][HA ] (Given) 1 [H 2 A]

HA- ¾¾ ® H+ + A--

\ K 2 = 5.0 ´ 10 -10 =

[H + ][A -- ] (Given) [HA - ]

1.0 ´10-4 ´ 283 ´100 gm /100ml 1000

= 2.83 × 10–3 gm/ 100 ml 29. (a) Let the initial moles of X be ‘a’ and that of Z be ‘b’ then for the given reactions, we

Initial At equi.

Given pKa = 4.5 and acid is 50% ionised. [HA] = [A–] (when acid is 50% ionised) \ pH = pKa + log 1 \ pH = pKa = 4.5 pOH = 14 – pH = 14 – 4.5 = 9.5 28. (c) Let s = solubility ˆˆ† Ag ∗ ∗ IO3, AgIO3 ‡ˆˆ s

or,

a moles a(1 – a)

K P1 =

(2aa)2 .PT1 [a(1 - a)][a(1 + a)]

 

Z

P + Q

Initial b moles 0 0 At equi. b(1 – a) ba ba (moles) Total no . of moles = b(1 – a) + ba + ba = b – ba + ba + ba = b(1 + a) Now K P = 2

or

s

Ksp = [Ag+] [IO3–] = s × s = s2 Given Ksp = 1 × 10–8

2Y

0 2aa (moles) Total no. of moles = a (1 – a) + 2aa = a – aa + 2aa = a (1 + a) Dn (n y ) 2 æ PT ö 1 ÷ ´ç Now, K P1 = ç ån ÷ nx è ø

= (1.0 × 10–5) × (5 × 10–10) = 5 × 10–15

éA- ù ë û pH = pK + log or a [ HA ]

   

X

have

[H + ]2 [A 2 - ] K= = K1 ´ K2 [H 2 A]

é salt ù 27. (d) For acidic buffer pH = pKa + log ê ú ë acid û

1´10-8

or

nQ ´ nP nz

é PT ù ´ê 2 ú ë ån û

Dn

(ba)(ba).PT2

K P2 =

[b(1 - a)][b(1 + a)]

K P1

4a 2 .PT1

K P2

=

(1 - a 2 )

´

(1 - a )2 PT2 .a 2

=

4PT1 PT2

C-39

Equilibrium

PT1

or

PT2 PT1

31.

é K P1 1 ù = given ú êQ êë K P2 9 úû

1 9

1 PT2 36 or 1 : 36 i.e., (a) is the correct answer. (c) Reaction (c) can be obtained by adding reactions (a) and (b) therefore K3 = K1. K2 Hence (c) is the correct answer. (c) The correct order of acidic strength of the given species in

or

30.

=

(iv)

32.

K1 =

(iii)

(ii)

x = [H+] = [ HCO3- ] = 1.195 × 10–4 Now, HCO3– (aq ) ∗ H 2O(l ) ƒ CO32– ( aq ) ∗ H3O∗ ( aq ) x, y

(i)

(a)

K2 =

1 ´10-4 M

So, K 2 ;

34.

5.1 ´ 10 -9 1 ´ 10 -4 base2

base1

® HPO4-- + H 3O + (ii) H 2PO4 + H 2O ¾¾ acid1

(iii)

base 2

base1

acid2

® H3PO4 + O -H 2PO 4- + OH - ¾¾ base1

acid 2

acid1

base2

Hence only in (ii) reaction H2PO4– is acting as an acid. 35.

(c)

H 2 CO3 (aq )∗ H2 O(l ) ƒ HCO3– (aq) ∗ H3O+ (aq) 0.034, x

x

y≥ x =y x

[H + ] = [ HCO3- ] = 1.195 × 10 –4 &

= 5.1 ´ 10 -5 M

acid 2

y ´ (x + y)

So the concentration of [H+] ; [ HCO3– ]= concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. Thus the final concentrations are

1´10-4 M

® H3O + + H 2 PO4(a) (i) H3PO4 + H 2O4 ¾¾ acid1

=

Þ K 2 = 4.8 ´ 10-11 = y = [CO32 - ]

K SP(BaCO3 ) = [Ba 2+ ][CO 32– ]

[Ba 2+ ] =

[CO32– ][H3O+ ]

Note from second step (y) = (x + y) [As x > > y so x + y ; x and x – y ; x]

2Na + + CO32-

1´10-4 M

y

( x - y) [HCO3– ] + + : [H3O ] = H from first step (x) and

1 (14 + 4.80 - 4.78) = 7.01 2

Na 2 CO3 ¾¾ ®

y

As HCO3- is again a weak acid (weaker than H2CO3) with x >> y.

1 1 1 pH = pKw + pKa - pKb 2 2 2 substituting given values, we get

33.

x´ x 0.034 - x

Þ 4.2 ≥10,7 ;

or (i) < (iii) < (ii) < (iv) It corresponds to choice (c) which is correct answer. (c) In aqueous solution BA(salt) hydrolyses to give BA + H2O BOH + HA Base acid Now pH is given by

pH =

[H 2CO3 ]

=

x2 Þ x < 1.195≥10,4 0.034 As H 2 CO 3 is a weak acid so the concentration of H2CO3 will remain 0.034 as 0.034 >> x.

=

HSO3 F > H 3 O + > HSO 4 - > HCO3-

[HCO3- ][H3O + ]

x

[CO23 - ] = 4.8 ´ 10 -11

36. (b)

ˆˆ† Ag∗ ∗ Br, AgBr ‡ˆˆ Ksp = [Ag+] [Br–] For precipitation to occur Ionic product > Solubility product [Br, ]
(CH3 ) 2 CH

- C – C – C– C– C – C– C – C – C – C –

CH3

CH3 . 2, 3-dichloro butane will

H H exhibit optical isomerism due to the presence of two asymmetric carbon atom.

d–

F

H O–H

d+

Due to hydrogen bonding between H & F gauche conformation is most stable hence the correct order is Eclipse, Anti, Gauche

C-59

Organic Chemistry-Some Basic Principles & Techniques

30.

(a)

CH3 7 6 5 4| 3 2 1 CH3 - CH 2 - CH 2 - C - CH - CH 2 - CH3 | | CH3 CH 2 | CH3

O

CH3 | 3 H3C- C - CH3 | CH3

37. (a)

1

NOTE The organic compounds which

(b)

have chiral carbon atom (a carbon atom attached to four different group or atoms and do not have plane of symmetry rotate plane polarised light. CHO | HO - C*- H (* is asymmetric carbon) | CH 2 OH

32. 33.

(a) Nitro group is electron withdrawing group, so it deactivates the ring towards electrophilic substitution. (b) Chiral conformation will not have plane of symmetry. Since twist boat does not have plane of symmetry it is chiral.

3

38. (b) –

34.

35.

36.

(b) The absolute configuration is (R, R) (using priority rules to get the absolute configuration) So the correct answer is (b) (b) In option (b) the complex formed is with benzene where as in other cases it is formed with nitrobenzene with –NO2 group in different position (o-, m-, p-). The complex formed with nitrobenzene in any position of –NO2 group is less stable than the complex formed with benzene so the correct answer is (b) NOTE The most stable complex has lowest energy. (a) The correct order of priority for the given functional group is +111.5º

– – Cl > C6H5CH2 > (CH3)2 CH

Cl

39. (b)

–M effect delocalises –ve charge

H3C

CH3

H

40. (b)

–

> (CH3)3C

+I effect of CH3 group intensifies the –ve charge

H3C

C=C

6

2

Cl C

–ve charge highly dispersed due to – I effect

41. 5

2

Neopentane or 2, 2- Dimethylpropane

1

4

||

–COOH > –SO3 H > – C - NH 2 > - C - H

3- ethyl - 4,4 -dimethyl heptane

31.

O

||

H C=C

H

H

CH3

*

CH 3 – CH = CH – CHCH 3 |

OH exhibits both geometrical as well as optical isomerism. cis - R cis - S trans - R trans - S (d) The correct order of basicity is RCOO - < CH º C - < NH 2 - < R -

42. (c) For a compound to show optical isomerism, presence of chiral carbon atom is a necessary condition. H | H 2 C < HC — C* — CH 2 , CH3 | CH3 3- methyl-1-pentene

43. (c) When either of the two forms of glucose is dissolved in water there is change in rotation till the equilibrium value of + 52.5º. This is known as mutarotation a–D(+)Glucose Equilibrium Mixture +111.5º

+52.5º

b–D–(+)Glucose +19.5º

EBD_7764 C-60

Chemistry

H

44. (d)

H

sp3 Carbon

Cyclopentadiene does not obey Huckel's Rule, as it has sp3 carbon in the ring. (c) Carbocations are planar hence can be attacked on either side to form racemic mixture.

45.

Å

SbCl

5 ® Ph - C H - CH + SbCl- ¾¾ Cl - CH- CH3 ¾¾¾¾ ® 3 6 Toluene | (carbocation) Ph (- )

Ph - CH - CH3 + SbCl5 | Cl (d+l) mixture

46.

(d) Higher stability of allyl and aryl substituted methyl carbocation is due to dispersal of positive charge due to resonance +

+

CH 2 = CH - C H 2 ¬¾® CH 2 - CH = CH 2 Resonating structures of allyl carbocation

+

CH2

CH2

CH2

H | 48. (c) H3C — C = CH — CH 2 | Ph 1- Phenyl-2-butene the two groups around each of the doubly bonded carbon Because, all are different. This compound can show cis-and trans-isomerism. 49. (c) Mass of substance = 250 mg = 0.250 g Mass of AgBr = 141 mg = 0.141 g 1 mole of AgBr = 1 g atom of Br 188 g of AgBr = 80 g of Br \ 188 g of AgBr contain bromine = 80 g 0.141 g of AgBr contain bromine = 80 ´ 0.141 188 This much amount of bromine present in 0.250 g of organic compound \ % of bromine = = 24% CO2H 50. (d) 1 OH H

Å

2

H

CH2

CH3

Å

At (1),

Å

Cl

3

whereas in alkyl carbocations dispersal of positive charge on different hydrogen atoms is due to hyperconjugation. Hence the correct order of stability will be

1

S

Resonating structures of benzyl carbocation

4

1

2

3

2

Å

4

It is ‘S’configurated At. (2),

CH2 Å

Å

2

> CH 2 = CH - C H 2 > CH3 - CH 2 - CH 2 Allyl, I

Benzyl, III

Propyl, II

4

1 1

3

R 2

H

47.

O

(d)

O

H

3

H S

51.

(d)

S

H In both the molecules the bond moments are not canceling with each other and hence the molecules has a resultant dipole and hence the molecule is polar.

4

Region 2 (blue flame) will be the hottest region of Bunsen flame shown in given figure

O

52. (d)

is nonaromatic and hence least reasonance stabilized whereas other three are aromatic.

C-61

Hydrocarbons

13

Hydrocarbons 1.

2.

3.

4.

5.

6.

7.

Which of these will not react with acetylene? [2002] (a) NaOH (b) ammonical AgNO3 (c) Na (d) HCl. What is the product when acetylene reacts with hypochlorous acid? [2002] (a) CH3COCl (b) ClCH2CHO (c) Cl2CHCHO (d) ClCH2COOH. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. This alkane could be [2003] (a) pentane (b) isopentane (c) neopentane (d) propane Butene-1 may be converted to butane by reaction with [2003] (a) Sn – HCl (b) Zn – Hg (c) Pd/H2 (d) Zn – HCl Which one of the following has the minimum boiling point? [2004] (a) 1 - Butene (b) 1 - Butyne (c) n- Butane (d) isobutane 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly [2005] (a) 1-bromo-3-methylbutane (b) 2-bromo-3-methylbutane (c) 2-bromo-2-methylbutane (d) 1-bromo-2-methylbutane Reaction of one molecule of HBr with one molecule of 1, 3-butadiene at 40°C gives predominantly [2005] (a) 1-bromo-2-butene under kinetically controlled conditions (b) 3-bromobutene under thermodynamically controlled conditions (c) 1-bromo-2-butene under thermodynamically controlled conditions (d) 3-bromobutene under kinetically controlled conditions

8.

Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is [2005] (a) 2-methylpentane (b) 2, 2-dimethylbutane (c) 2, 3-dimethylbutane (d) n-hexane 9. The compound formed as a result of oxidation of ethyl benzene by KMnO4 is [2007] (a) benzyl alcohol (b) benzophenone (c) acetophenone (d) benzoic acid. 10. Which of the following reactions will yield 2, 2-dibromopropane? [2007] (a) CH3 – CH = CH2 + HBr ® (b) CH3 – C º CH + 2HBr ® (c) CH3CH = CHBr + HBr ® (d) CH º CH + 2HBr ® 11. The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly [2007] (a) m-chlorobenzene (b) benzoyl chloride (c) benzyl chloride (d) o- and p-chlorotoluene. 12. Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated wth cuprous bromide. The reaction mixture so formed contains [2008] (a) mixture of o- and p-bromotoluenes (b) mixture of o- and p-dibromobenzenes (c) mixture of o- and p-bromoanilines (d) mixture of o- and m-bromotoluenes 13. In the following sequence of reactions, the alkene affords the compound ‘B’ O

H O Zn

2 ® B. CH 3 - CH = CH - CH 3 ¾¾3¾ ® A ¾¾¾

The compound B is

[2008]

EBD_7764 C-62

14.

15.

Chemistry

(a) C H3CH2CHO (b) C H3COCH3 (c) C H3CH2COCH3 (d) C H3CHO The hydrocarbon which can react with sodium in liquid ammonia is [2008] (a)

CH 3CH 2 CH 2 C º CCH 2CH 2 CH3

(b)

CH 3CH 2C º CH

(c)

CH 3CH = CHCH3

(d)

CH 3CH 2C º CCH 2CH3

(a)

CH 3 - CH = CH 2

(b)

CH 3C º C - CH3 H H | | CH 3 - C = C - CH3

(c)

17.

18.

19.

20.

CH3 (b) H3C

(a) CH3

CH3

CH3 CH3

(c)

(d)

CH3

The treatment of CH3MgX with CH 3C º C - H produces

16.

CH3

[2008]

(d) CH4 One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is [2010] (a) propene (b) 1-butene (c) 2-butene (d) ethene Ozonolysis of an organic compound 'A' produces acetone and propionaldhyde in equimolar mixture. Identify 'A' from the following compounds: [2011RS] (a) 1 – Pentene (b) 2 – Pentene (c) 2 – Methyl – 2 – pentene (d) 2 – Methyl – 1 – pentene Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alkyl halide ? [2012] (a) Tertiary butyl chloride (b) Neopentane (c) Isohexane (d) Neohexane 2-Hexyne gives trans-2-Hexene on treatment with : [2012] (a) Pt/H2 (b) Li / NH3 (c) Pd/BaSO4 (d) Li AlH4 Which compound would give 5 - keto - 2 methylhexanal upon ozonolysis ? [JEE M 2015]

21. The product of the reaction given below is: 1. NBS/hv 2. H 2O/K 2CO3

X

[JEE M 2016]

CO2H

O (a)

(b)

OH (c)

(d)

22. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate: [JEE M 2016] (a)

CH 3 – CH ( OH ) - CH 2+

(b)

CH3 – CHCl - CH 2+

(c)

CH3 – CH + - CH 2 – OH

(d)

CH3 – CH + - CH 2 – Cl

23. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is : [JEE M 2017] (a) Six

(b) Zero

(c) Two

(d) Four

C-63

Hydrocarbons

Answer Key 1 (a) 16 (c)

1.

2 (c) 17 (c)

3 (c) 18 (b)

4 (c) 19 (b)

5 (d) 20 (d)

6 7 8 (c) (c) (c) 21 22 23 (d) (d) (d)

9 (d)

+HCl

CHCl

CH º CH

Na liq. NH 3

CH º CH

+HCl

(c)

13 (d)

alkynes > alkene > alkanes (straight chain) > branched chain alkanes.

CHCl2 [AgNO3+NH4OH]

CH3

6.

(c)

|

| |

Br

2-bromo-2-methylbutane

Ease of replacement of H-atom 3° > 2° > 1°. (c) CH2 = CH – CH = CH2CH2 + HBr Br

CH2 = CH – CH – CH3

dichloroacetaldehyde

At –80°C the product is 1, 2-addition

(c) In neopentane all the H atoms are same (1º).

CH2 – CH = CH – CH3

CH3 | CH 3 - C - CH 3 | CH3

Butene - 1 ¾¾¾¾® Butane 5.

(d)

NOTE Among isomeric alkanes, the

straight chain isomer has higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. Further due to the presence of p electrons, these moleculs are slightly polar and hence have higher

sun light

CH3 - C - CH2CH3

CHOH CH º CH + HOCl ¾¾ ® || CHCl

H 2 / Pd

2 ¾¾® ¾

CH3

AgCº CAg + NH4NO3

(c) Alkenes combine with hydrogen under pressure and in presence of a catalyst (Ni, Pt or Pd) and form alkanes.

Br

CH3 - CH - CH2 - CH3

CH (OH ) 2 ù CHO - H 2O HOCl é ¾¾¾®ê | ¾® | ú ¾¾ ¾ CHCl 2 ëê CHCl 2 ûú

4.

15 (d)

CHCl

7.

3.

14 (b)

Thus B.pt. follows the order

white ppt.

2.

12 (a)

CH 2

CH3

[AgNO +NH OH]

11 (d)

boiling points than the corrosponding alkanes.

(a) Acetylene reacts with the other three as: CH º CNa

10 (b)

Br

At 40°C the product is 1, 4-addition

CH3 CH3

8.

|

|

(c) CH 3 - C H - C H - CH 3 . Since it contains only two types of H-atoms hence it will give only two mono chlorinated CH3 CH3 |

|

compounds viz. Cl.CH 2 - C H - C H - CH 3 1-chloro -2,3 -dimethyl butane

CH 3 CH 3 |

and

|

CH 3 - C - C H - CH 3 |

Cl

2 -chloro - 2,3-dimethyl butane

EBD_7764 C-64 9. (d) When alkyl benzene are oxidised with alkaline KMnO4, (strong oxidising agent) the entire alkyl group is oxidised to –COOH group regardless of length of side chain.

Chemistry

CH3 (HNO3+ HSO) 2

4

CH2CH3

( O ), KMnO4 / OH¾¾¾¾¾¾¾¾ ® Ethyl benzene

CH3

(HNO + HSO)

COOH

CH3

NO2 +

NO2 p-

o-

Benzoic aicd

10. (b) The reaction follows Markownikoff rule which states that when unsymmetrical reagent adds across unsymmetrical double or triple bond the negative part adds to carbon atom having lesser number of hydrogen atoms.

on reduction with Sn/HCl they will form corresponding anilines in which –NO2 group changes to –NH2. The mixture now contains CH3

CH3 - C º CH + HBr ®

2, 2dibromo-propane

11. (d) FeCl3 is Lewis acid. In presence of FeCl3 side chain hydrogen atoms of toluene are substituted.

CH3

Toluene

. These anilines

when diazotized and then treated with CuBr forms o-, p- bromotoluenes. 13. (d) Completing the sequence of given reactions, O

CH 3 – CH = CH - CH 3 ¾¾3¾ ®

O Zn / H O

2 CH – CH3 ¾¾¾¾®

CH3– CH

Cl

FeCl3 + Cl2 ¾¾¾®

and NH2

Br | ® CH3 - C - CH3 CH3 - C = CH 2 ¾¾¾ | | Br Br HBr

CH3

CH3 NH2

+

O

O ‘A’ (ozonide)

o-chloro toluene

CH3

2CH3CHO+ H 2 O + ZnO 'B '

Cl

p-chloro toluene

CH3 12. (a)

NOTE Toluene (

) contains –

CH3 group which is o-, p- directing group so on nitration of toluene the –NO2 group will occupy o-, p- positions.

Thus ‘B’ is CH3CHO Hence (d) is correct answer. 14. (b) Alkynes having terminal –C º H react with Na in liquid ammonia to yield H2 gas of the given compounds CH3CH2C º CH can react with Na in liquid NH3 so the correct answer is (b). Na in liquid NH3

CH3CH 2 C º CH ¾¾¾¾¾ ®

Hydrocarbons

1 CH3 CH 2 C º C – Na + + H 2 (g) 2 15. (d) Writing the reaction we get CH 3 MgX + CH 3 – C º C – H ¾¾ ®

CH3 – C º CMgX + CH 4 (g) So we find that CH4 is produced in this reaction. 16. (c) The given molecular formula suggests that the aldehyde formed will be acetaldehyde hence the alkene will be

C-65 19. (b) Anti addition of hydrogen atoms to the triple bond occurs when alkynes are reduced with sodium (or lithium) metal in ammonia, ethylamine, or alcohol at low temperatures. This reaction called, a dissolving metal reduction, produces an (E)- or trans-alkene. Sodium in liq. NH3 is used as a source of electrons in the reduction of an alkyne to a trans alkene.

CH3

CH2

2-Hexyne

Li/NH 3

CH3

Birch reduction

CH3CH = CH CH3

CH2

C

CH2

C

CH2

H

C

C

H

2- butene

CH3

Trans-2-Hexene

O3 H ¾¾®

O C

C

O

O

20.

H

(d)

When 1, 3-dimethylcyclopentene is heated with ozone and then with zinc and acetic acid, oxidative cleavage leads to keto - aldehyde.

CH3

17.

CH3

1O - 78° C 2 - Zn- CH 3 COOH

¾ ¾ ¾3¾ ¾ ¾ ¾ ®

Zn / H 2O 2CH3CHO + H 2O 2 ¾¾¾¾® (c) From the products formed it is clear that the compound has 5 carbon atoms with a double bond and methyl group on 2 nd carbon atom.

CH3

O O C–H

CH3

O O || || CH3 — C— CH 2 — CH 2 — CH— C— H | CH3

CH3 | CH3 - C = CH - CH 2 - CH3 O / Zn, H O

3 2 ® ¾¾¾¾¾¾ (2–Methyl–2–pentene) (A)

CH3

21.

(d)

5- keto – 2 – methylhexanal N – bromosuccinimide results into bromination at allylic and benzylic positions

CH3 O | CH3 - C = O + CH3 - CH 2 - C

NBS/hv

H

Acetone

Propionaldehyde

18. (b)

More stable

CH3 H3C

C CH3

CH3

neopentane

Cl2/hv monohalogenation

Br

single product

NBS

H 2O/K 2CO 3

HO

EBD_7764 C-66

22.

Chemistry

(d)

CH2 – CH – CH3 Cl More stable intermediate

Å

CH2 = CH – CH3 + Cl –

Å

CH2 – CH – CH3 Cl

CH3 H

Br

H

CH3

Å

CH2 – CH – CH3 OH CH2 – CH – CH3 Cl Cl OH 23. (d) If two chirality centres are created as a result of an addition reaction four stereoisomers can be obtained as products.

CH3 H

C = C

CH3 C2H5

HBr ¾¾ ¾® – H2 O

cis-3, methyl pent-2-ene

Br CH3 | | CH3 – CH – CH – CH2 – CH3 * * 2, Bromo, 3-methyl pentane (2 chiral centre)

CH3 Br

H

H3C

H

C2H5

C2H5

(I)

(II)

CH3

CH3

Br

H

H

CH3 C2H5 (III)

H

Br

H3C

H C2H5 (IV)

C-67

Environmental Chemistry

14

Environmental Chemistry 1.

2.

3.

The smog is essentially caused by the presence of [2004] (a) Oxides of sulphur and nitrogen (b) O2 and N2 (c) O2 and O3 (d) O3 and N2 Identify the wrong statement in the following: [2008] (a) Chlorofluorocarbons are responsible for ozone layer depletion (b) Greenhouse effect is responsible for global warming (c) Ozone layer does not permit infrared radiation from the sun to reach the earth (d) Acid rain is mostly because of oxides of nitrogen and sulphur Identify the incorrect statement from the following : [2011RS] (a) Ozone absorbs the intense ultraviolet radiation of the sun. (b) Depletion of ozone layer is because of its chemical reactions with chlorofluoro alkanes.

4.

5.

6.

(c) Ozone absorbs infrared radiation. (d) Oxides of nitrogen in the atmosphere can cause the depletion of ozone layer. What is DDT among the following ? [2012] (a) Greenhouse gas (b) A fertilizer (c) Biodegradable pollutant (d) Non-biodegradable pollutant The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was: [2013] (a) Methyl isocyanate (b) Methylamine (c) Ammonia (d) Phosgene A water sample has ppm level concentration of following anions [2017] – 2– – F = 10; SO4 = 100; NO3 = 50 the anion/anions that make/makes the water sample unsuitable for drinking is/are : (b) both SO42– and NO3– (a) only NO3– – (c) only F (d) only SO42–

Answer Key 1 (a)

1. 2.

3.

2 (c)

3 (c)

4 (d)

5 (a)

6 (c)

(a) Photochemical smog is caused by oxides of sulphur and nitrogen. NOTE Ozone layer acts as a shield and (c) does not allow ultraviolet radiation from sun to reach earth. It does not prevent infrared radiation from sun to reach earth. Thus option (c) is wrong statement and so it is the correct answer. (c) The ozone layer, existing between 20 to 35 km above the earth’s surface, shield the earth from the harmful U. V. radiations from the sun. Depletion of ozone is caused by oxides of nitrogen

4. 5. 6.

N2O + h u ¾ ¾® NO + N reactive nitric oxide NO + O 3 ¾ ¾® NO 2 + O 2 O3 + h u ¾ ¾® O 2 + O ¾® NO + O 2 NO 2 + O ¾ 2 O3 + h u ¾ ¾® 3 O 2 (Net reaction) The presence of oxides of nitrogen increase the decomposition of O 3 . (d) DDT is a non-biodegradable pollutant. (a) (c) Above 2 ppm concentration of F – in drinking water cause brown mottling of teeth.

EBD_7764 C-68

Chemistry

15

The Solid State 1.

2.

3.

4.

5.

Na and Mg crystallize in BCC and FCC type crystals respectively, then the number of atoms of Na and Mg present in the unit cell of their respective crystal is [2002] (a) 4 and 2 (b) 9 and 14 (c) 14 and 9 (d) 2 and 4. How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g ? [2003] [Atomic masses : Na = 23, Cl = 35.5] (a) 5.14 × 1021 unit cells (b) 1.28 × 1021 unit cells (c) 1.71 × 1021 unit cells (d) 2.57 × 1021 unit cells What type of crystal defect is indicated in the diagram below ? [2004] Na + Cl - Na + Cl - Na + Cl Cl– c Cl– Na+ c Na+ Na+Cl– c Cl– Na+ Cl– Cl– Na+ Cl–Na+ c Na+ (a) Interstitial defect (b) Schottky defect (c) Frenkel defect (d) Frenkel and Schottky defects An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be [2005]

(a)

A 3B

(b) AB3

(c)

A 2B

(d) AB

Total volume of atoms present in a face-centred cubic unit cell of a metal is (r is atomic radius) [2006] (a)

12 3 pr 3

(b)

16 3 pr 3

(c) 6.

20 3 pr 3

(d)

24 3 pr 3

In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be [2008] (a) X4 Y3 (b) X2 Y3 (c) X2 Y (d) X3 Y4 7. Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom? [2009] (a) 127 pm (b) 157 pm (c) 181 pm (d) 108 pm 8. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is (a) 288 pm (b) 398 pm [2010] (c) 618 pm (d) 144 pm 9. Percentages of free space in cubic close packed structure and in body centered packed structure are respectively [2010] (a) 30% and 26% (b) 26% and 32% (c) 32% and 48% (d) 48% and 26% 10. Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is : [2011RS] (a) 108 pm (b) 128 pm (c) 157 pm (d) 181 pm 11. Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be : [2012] (a) 75 pm (b) 300 pm (c) 240 pm (d) 152 pm 12. Which of the following exists as covalent crystals in the solid state ? [2013] (a) Iodine (b) Silicon (c) Sulphur (d) Phosphorus

The Solid State 13. CsCl crystallises in body centered cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct? [2014]

14.

(a)

r

+r

= 3a

(b)

r

+r

=

(c)

r

+r

=

(d)

r

+r

= 3a

Cs + Cs +

Cs + Cs +

Cl Cl -

Cl -

Cl -

3a 2

3 a 2

C-69 (c) It contains Cs3+ and I– ions. (d) It contains Cs+, I– and lattice I2 molecule. 15. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29Å. The radius of sodium atom is approximately : [JEE M 2015] (a) 5.72Å (b) 0.93Å (c) 1.86Å (d) 3.22Å 16. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be : [JEE M 2017]

The correct statement for the molecule, CsI3 is: (a) It is a covalent molecule. [2014] (b) It contains Cs+ and I3- ions.

(a) 2a

(b) 2 2 a

(c)

(d)

2a

a 2

Answer Key 1 (d) 16 (d)

1.

2 (d)

3 (b)

4 (b)

5 (b)

6 (a)

7 (a)

(d) In bcc - points are at corners and one in the centre of the unit cell. Number of atoms per unit cell 1 = 8´ +1 = 2 . 8 In fcc - points are at the corners and also centre of the six faces of each cell. Number of atoms per unit cell

2.

8 (d)

1 1 = 8´ + 6 ´ = 4 . 8 2 (d) Since in NaCl type of structure 4 formula units form a cell. Number of formulas in

cube shaped crystals =

1. 0 ´ 6.02 ´ 10 23 58.5

No. of unit cells present in a cubic crystal 3

=

P ´ a ´ NA m ´ NA = M´Z M´Z

9 (b)

10 (b)

11 (d)

\ units cells = 3.

4.

12 (b)

13 (c)

14 (b)

15 (c)

1.0 ´ 6.02 ´10 23 = 2.57 × 1021 58.5 ´ 4

unit cells. When equal number of cations and anions are missing from their regular lattice positions, we have schottky defect. This type of defects are more common in ionic compounds with high co-ordination number and where the size of positive and negative ions are almost equal e.g. NaCl KCl etc. (b) Number of A ions in the unit cell.

(b)

1 ´8 = 1 8 Number of B ions in the unit cell

=

1 ´6 = 3 2 Hence empirical formula of the compound = AB3

=

EBD_7764 C-70 5. (b) The face centered cubic unit cell contains 4 atom

6.

4 16 \ Total volume of atoms = 4 ´ pr 3 = pr 3 3 3 (a) From the given data, we have Number of Y atoms in a unit cell = 4 Number of X atoms in a unit cell

Chemistry

11.

(d) For BCC structure

3 a = 4r

3 3 a = ´ 351 = 152 pm. 4 4

r=

12. (b) 13. (c)

2 16 = 8´ = 3 3

Cl –

Cl – Cl –

Cl

From the above we get the formula of the Cs+

compound as X16 / 3Y4 or X 4 Y3 7.

(a) For fcc unit cell, 4r =

8.

2 ´ 361 = 127 pm r= 4 (d) For an Fcc crystal rcation + ranion

2a

a 2 = 4r

r=

361 ´ 2 = 127.6 » 128pm 4

Cl

Cl –

–

Relation between radius of cation, anion and edge length of the cube

edge length = 2

508 110 + ranion = 2 ranion = 254 – 110 = 144 pm 9. (b) Packing fraction is defined as the ratio of the volume of the unit cell that is occupied by the spheres to the volume of the unit cell. P.F. for ccp and bcc are 0.74 and 0.68 respectively. So, the free space in ccp and bcc are 26% & 32% respectively. 10. (b) fcc lattice a = 361 pm

Cl

Cl –

2r

Cs +

Cl-

= 3a

3a 2 (b) CsI3 dissociates as CsI3 ® Cs+ + I3– (c) In bcc the atoms touch along body diagonal r

+r

\

2r + 2r =

\

r=

Cs +

14. 15.

+ 2r

Cl-

=

3a

3a 3 ´ 4.29 = = 1.857Å 4 4

16. (d) For a FCC unit cell

r=

2a 4

\ closest distance (2r) =

2a a = 4 2

C-71

Solutions

16

Solutions 1.

2.

3.

4.

5.

Freezing point of an aqueous solution is (–0.186)°C. Elevation of boiling point of the same solution is Kb = 0.512°C,Kf = 1.86°C, find the increase in boiling point. [2002] (a) 0.186°C (b) 0.0512°C (c) 0.092°C (d) 0.2372°C. In mixture A and B components show -ve deviation as (a) D Vmix > 0 [2002] (b) D Hmix < 0 (c) A – B interaction is weaker than A – A and B – B interaction (d) A – B interaction is stronger than A – A and B – B interaction. If liquids A and B form an ideal solution [2003] (a) the entropy of mixing is zero (b) the free energy of mixing is zero (c) the free energy as well as the entropy of mixing are each zero (d) the enthalpy of mixing is zero In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking kf for water as 1.85, the freezing point of the solution will be nearest to [2003] (a) – 0.360º C (b) – 0.260º C (c) + 0.480º C (d) – 0.480º C A pressure cooker reduces cooking time for food because [2003] (a) boiling point of water involved in cooking is increased (b) the higher pressure inside the cooker crushes the food material (c) cooking involves chemical changes helped by a rise in temperature (d) heat is more evenly distributed in the cooking space

6.

7.

8.

9.

Which one of the following aqueous solutions will exihibit highest boiling point ? [2004] (a) 0.015 M urea (b) 0.01 M KNO3 (c) 0.01 M Na2SO4 (d) 0.015 M glucose For which of the following parameters the structural isomers C2 H5 OH and CH3 OCH3 would be expected to have the same values?(Assume ideal behaviour) [2004] (a) Boiling points (b) Vapour pressure at the same temperature (c) Heat of vaporization (d) Gaseous densities at the same temperature and pressure Which of the following liquid pairs shows a positive deviation from Raoult’s law ? [2004] (a) Water - nitric acid (b) Benzene - methanol (c) Water - hydrochloric acid (d) Acetone - chloroform Which one of the following statements is FALSE? [2004] (a) The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl 2 > KCl > CH 3COOH > sucrose

(b) The osmotic pressure (p) of a solution is given by the equation p = MRT, where M is the molarity of the solution (c) Raoult’s law states that the vapour pressure of a component over a solution is proportional to its mole fraction (d) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression 10. Benzene and toluene form nearly ideal solution. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial

EBD_7764 C-72

11.

12.

13.

14.

15.

vapour pressure of benzene at 20°C for a solution containing 78 g of benzene and 46 g of toluene in torr is [2005] (a) 53.5 (b) 37.5 (c) 25 (d) 50 Equimolar solutions in the same solvent have [2005] (a) Different boiling and different freezing points (b) Same boiling and same freezing points (c) Same freezing point but different boiling points (d) Same boiling point but different freezing points Among the following mixtures, dipole-dipole as the major interaction, is present in [2006] (a) KCl and water (b) benzene and carbon tetrachloride (c) benzene and ethanol (d) acetonitrile and acetone 18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100ºC is [2006] (a) 76.00 Torr (b) 752.40 Torr (c) 759.00 Torr (d) 7.60 Torr A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be [2007] (a) 360 (b) 350 (c) 300 (d) 700 Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is [2007] (a) 1/2 (b) 2/3 (c)

16.

1 273 ´ 3 298

(d) 1/3.

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol–1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass of the substance will be [2007]

17.

18.

19.

20.

21.

Chemistry (a) 210.0 g mol–1 (b) 90.0 g mol–1 (c) 115.0 g mol–1 (d) 105.0 g mol–1. At 80° C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80° C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) [2008] (a) 52 mol percent (b) 34 mol percent (c) 48 mol percent (d) 50 mol percent The vapour pressure of water at 20° C is 17.5 mm Hg. If 18 g of glucose (C6 H12 O 6) is added to 178.2 g of water at 20° C, the vapour pressure of the resulting solution will be [2008] (a) 17.325 mm Hg (b) 15.750 mm Hg (c) 16.500 mm Hg (d) 17.500 mm Hg A binary liquid solution is prepared by mixing nheptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution? [2009] (a) The solution is non-ideal, showing – ve deviation from Raoult’s Law. (b) The solution is non-ideal, showing + ve deviation from Raoult’s Law. (c) n-heptane shows + ve deviation while ethanol shows – ve deviation from Raoult’s Law. (d) The solution formed is an ideal solution. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure ( in mmHg) of X and Y in their pure states will be, respectively: [2009] (a) 300 and 400 (b) 400 and 600 (c) 500 and 600 (d) 200 and 300 If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (DTf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) [2010] (a) 0.372 K (b) 0.0558 K (c) 0.0744 K (d) 0.0186 K

Solutions 22. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1) [2010] (a) 72.0 kPa (b) 36.1 kPa (c) 96.2 kPa (d) 144.5 kPa 23. A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/ mol is : [2011RS] (a) 171.2 (b) 68.4 (c) 34.2 (d) 136.2 24. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is : [2012] (a) 0.50 M (b) 1.78 M (c) 1.02 M (d) 2.05 M 25. Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC ? [2012] (a) 72 g (b) 93 g (c) 39 g (d) 27 g 26. The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be : [2013]

C-73 (a) 0.875 M (b) 1.00 M (c) 1.75 M (d) 0.975 M 27. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3 (PO4)2 (aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? [2014] (a) They all have the same osmotic pressure. (b) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (c) 0.125 M Na3 PO 4 (aq) has the highest osmotic pressure. (d) 0.500 M C2 H5OH(aq) has the highest osmotic pressure. 28. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is : [JEE M 2015] (a) 128 (b) 488 (c) 32 (d) 64

29. The freezing point of benzene decreases by 0.45°C when 0.2g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : [JEE M 2017] (Kf for benzene = 5.12 K kg mol –1) (a) 64.6% (b) 80.4% (c) 74.6% (d) 94.6%

Answer Key 1 (b) 16 (a)

1.

(b)

2 (d) 17 (d)

3 (d) 18 (a)

4 (d) 19 (b)

5 (a) 20 (b)

6 (c) 21 (b)

7 (d) 22 (a)

8 (b) 23 (b)

9 (d) 24 (d)

10 (d) 25 (b)

11 (d) 26 (a)

12 (d) 27 (a)

13 (b) 28 (d)

DTb = K b

WB ´1000 ; M B ´ WA

DTb K b DTb = = DTf -0.186 Kf

DTf = K f

WB ´1000 ; M B ´ WA

=

0.512 = 0.0512°C. 1.86

14 (b) 29 (d)

15 (d)

EBD_7764 C-74 2. (d) In solution containing A and B component showing negative deviation A–A and B–B interactions are weaker than that of A–B interactions. For such solutions. DH = –ve and DV = –ve 3. (d) When A and B form an ideal solution, DHmix = 0 4. (d) DTf = Kf × m × i ; DTf = 1.85 × 0.2 × 1.3 = 0.480º C \ Tf = 0 – 0.480ºC = – 0.480ºC

( HX

1- 0.3

5.

6.

+

H + X - , i = 1.3)

0. 3

0.3

NOTE On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases. (c) Q DTb° = Tb – Tbº Where Tb = b.pt of solution Tb° = b.pt of solvent or Tb = Tb° + DTb

(a)

NOTE Elevation in boiling point is a

colligative property, which depends upon the no. of particles.Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point.

7.

8.

Na2SO4 2Na + SO4 Since Na2SO4 has maximum number of particles (3) hence has maximum boiling point. (d) Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure. The heat of vaporisation, V.P. and b.pts will differ due to H-bonding in ethanol. (b)

NOTE Positive deviations are shown by such solutions in which solventsolvent and solute-solute interactions are stronger than the solvent interactions. In such solution, the intcractions among molecules becomes weaker. Therefore their escaping tendency increases which results in the increase in their partial vapour pressures. In a solutions of benzene and methanol there exists inter molecular H– bonding.

In this solution benzene molecules come

Chemistry between ethanol molecules which weaken intermolecular forces. This results in increase in vapour pressure.

DTf = K f ´ m ´ i . Since Kf has different values for different solvents, hence even if the m is the same DTf will be different 10. (d) Given, Vapour pressure of benzene = 75 torr Vapour pressure of benezene = 22 torr mass of benzene in = 78g 78 hence moles of benzene = = 1mole 78 (mol.wt of benzene = 78) mass of toluence in solution = 46g 46 hence moles of toluene = = 0.5 mole 92 now partial pressure of benezene 1 1 = 50 torr = 75 × = Pºb. Xb = 75 × 1 + 0.5 1.5 2 = 75 × = 50 3 11. (d) Equimolar solutions of normal solutes in the same solvent will have the same b. pts and same f. pts.

9.

(d)

d+

d-

12. (d) Acetonitrile ( CH 3 - C º N ) and acetone d+

(CH3) d–

d–

C = O both are polar

(CH3) molecules, hence dipole-dipole interaction exist between them. Between KCl and water ion-dipole interaction is found and in Benzene ethanol and Benzene–Carbon tetra chloride dispersion force is present 13. (b) Moles of glucose = 18 = 0.1 180 1 78 .2 Moles of water = = 9.9 18 Total moles = 0.1 + 9.9 = 10

p H 2O = Mole fraction × Total pressure 9.9 = ´ 760 10 = 752.4 Torr

C-75

Solutions

14.

(b)

PAo

or 760 = 520X A + 1000 - 1000X A

PBo

= ? , Given = 200mm , xA = 0.6, xB = 1 – 0.6 = 0.4, P = 290

or 480X A = 240

P = PA + PB = PAo x A + PBo x B

15.

Þ 290 = PAo × 0.6 + 200 × 0.4 \ PAo = 350 mm (d) Let the mass of methane and oxygen = m gm. Mole fraction of O2

=

240 1 = or 50 mol. percent 480 2 i.e., The correct answer is (d)

or X A =

NOTE On addition of glucose to water,,

18. (a)

vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation

Moles of O2 Moles of O 2 + Moles of CH 4

Po - Ps Moles of glucose in solution = Ps moles of water in solution

1 m / 32 m / 32 = = 3 m / 32 + m /16 3m / 32 Partial pressure of O2 = Total pressure ×

=

mole fraction of O2 , PO2 = P × 16.

1 1 = P 3 3

(a) Osmotic pressure ( p ) of isotonic solutions are equal. For solution of unknown substance (p = CRT) 5.25 / M V For solution of urea, C2 (concentration) = C1 =

1.5 / 60 V Given, p1 = p2 Q p = CRT \ C1RT = C2RT or C1 = C2

5.25 / M 1.8 / 60 = V V M = 210 g/mol \ (d) At 1 atmospheric pressure the boiling point of mixture is 80°C. At boiling point the vapour pressure of mixture, PT = 1 atmosphere = 760 mm Hg. Using the relation,

or

17.

PT = PAo X A + PBo X B , we get

PT = 520X A + 1000(1 - X A ) {Q PAo = 520mm Hg , PBo

= 1000 mm Hg , X A + X B = 1 }

or or

17.5 - Ps 18/180 = [Q Po = 17.5 ] Ps 178.2/18

17.5 – Ps = 0.1 ´ Ps or Ps = 17.325 mm Hg.

9.9 Hence (a) is correct answer. 19. (b) For this solution intermolecular interactions between n-heptane and ethanol aare weaker than n-heptane - n-heptane & ethanol-ethanol interactions hence the solution of n-heptane and ethanol is nonideal and shows positive deviation from Raoult’s law.

20. (b)

Ptotal = PA° X A + PB° X B

1 3 550 = PA° ´ + PB° ´ 4 4 PA° + 3PB° = 550 ´ 4 In second case Ptotal = PA° ´

...(i)

1 4 + PB° ´ 5 5

PA° + 4PB° = 560 ´ 5 Subtract (i) from (ii) \ PB° = 560 ´ 5 - 550 ´ 4 = 600

Q PA° = 400 21. (b) Sodium sulphate dissociates as Na 2SO 4 (s) ¾¾ ® 2Na + + SO -4 hence van’t hoff factor i = 3 Now D T f = i k f .m = 3 × 1.86 × 0.01 = 0.0558 K

...(ii)

EBD_7764 C-76

22.

Chemistry

(a)

PTotal = = P° A x A + P° B X B = P°Heptane X Heptane + P°Octane X Octane 25 /100 35 /114 = 105 ´ 25 35 + 45 ´ 25 35 + + 100 114 100 114 0.25 0.3 + 45 ´ = 105 ´ 0.25 + 0.3 0.25 + 0.3

105 ´ 0.25 45 ´ 0.3 26.25 + 13.5 + = 0.55 0.55 0.55 = 72 kPa (b) For isotonic solutions

=

23.

p1 = p 2 C1 = C2

Þ

p

C2 H5 OH = 1´ 0.500 ´ R ´ T = 0.5 RT

p

Mg3 (PO 4 )2

= 5 ´ 0.100 ´ R ´ T = 0.5 RT pKBr

= 2 ´ 0.250 ´ R ´ T = 0.5 RT

p Na PO 3 4

= 4 ´ 0.125 ´ RT = 0.5 RT

Since the osmotic pressure of all the given solutions is equal. Hence all are isotonic solution. 28. (d) Using relation,

5 / 342 1/ M = 0.1 0.1

p° - ps w 2 M1 = ps w1M 2

5 1 = 342 M

where w1, M1 = mass in g and mol. mass of solvent w2, M2 = mass in g and mol. mass of solute Let M2 = x p° = 185 torr ps = 183 torr

M =

342 = 68.4 gm/mol 5

24. (d) Molarity =

moles of solute volume of solution(l)

Mass of solution = 1000 + 120 = 1120 M M 1120 d = ;v = = mL v d 1.15 120 ´ 1.15 = ´ 1000 = 2.05 M 60 ´ 1120

25. (b) DTf = i × Kf × m Given DTf = 2.8, Kf = 1.86 K kg mol–1 i = 1 (ethylene glygol is a non- electrolyte) wt. of solvent = 1 kg Let of wt of solute = x Mol. wt of ethylene glycol = 62 2.8 = 1 × 1.86 ×

x 62 ´ 1

2.8 ´ 62 = 93 gm 1.86 (a) From molarity equation : M1V1 + M2V2 = M × V

or x =

26.

27.

M V + M 2V2 M= 1 1 where V = total volume V 750 ´ 0.5 + 250 ´ 2 = 1000 = 0.875 M (a) p = i CRT

185 - 183 = 183

(Mol. mass of acetone = 58)

x = 64 \ Molar mass of substance = 64 29. (d) In benzene 2CH3COOH ƒ (CH3COOH)2 1–a a/2 i = 1 – a + a/2 = 1 – a/2 Here a is degree of association DTf = iKfm æ 0.2 ö çè ÷ æ aö 60 ø 0.45 = ç1 – ÷ (5.12) è 20 2ø 1000 a = 0.527 2 a = 0.945 % degree of association = 94.6% 1–

C-77

Electrochemistry

17

Electrochemistry 1.

2.

3.

Conductivity (unit Siemen’s S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is (a) Sm mol–1 (b) Sm2 mol–1 [2002] (c) S–2m2 mol (d) S2m2 mol–2. EMF of a cell in terms of reduction potential of its left and right electrodes is [2002] (a) E = Eleft - Eright (b) E = Eleft + Eright (c) E = Eright - Eleft (d) E = – (Eright + Eleft). What will be the emf for the given cell [2002] Pt | H2 (P1) | H+ (aq) | | H2 (P2) | Pt (a) (c)

4.

5.

6.

P RT log e 1 F P2 P RT log e 2 F P1

P2 RT (b) 2 F loge P 2

(d) none of these.

Which of the following reaction is possible at anode? [2002] 3+ 2– + (a) 2 Cr + 7H2O ® Cr2O7 + 14H (b) F2 ® 2F –

(c) (1/2) O2 + 2H+ ® H2O (d) none of these. When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are [2002] Cathode Anode (a) pure zinc pure copper (b) impure sample pure copper (c) impure zinc impure sample (d) pure copper impure sample. For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25ºC will be [2003]

7.

8.

9.

(a) 29.5 × 10–2 (b) 10 (c) 1 × 1010 (d) 1 × 10–10 Standard reduction electrode potentials of three metals A, B & C are respectively + 0.5 V, – 3.0 V & –1.2 V. The reducing powers of these metals are [2003] (a) A > B > C (b) C > B > A (c) A > C > B (d) B > C > A When during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be [2003] (a) 10.8 g (b) 21.6 g (c) 108 g (d) 1.08 g For the redox reaction : [2003] Zn (s) + Cu 2 + (0.1 M ) ® Zn 2 + (1 M ) + Cu (s) º taking place in a cell, E cell is 1.10 volt. Ecell for

RT ö the cell will be æç 2.303 = 0.0591÷ [2003] F è ø (a) 1.80 volt (b) 1.07 volt (c) 0.82 volt (d) 2.14 volt 10. In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to [2004] (a) produce high purity water (b) create potential difference between two electrodes (c) generate heat (d) remove adsorbed oxygen from elctrode surfaces 11. Consider the following Eº values

Eº

Fe3 + / Fe 2 +

= +0.77 V ; E º

Sn 2 + / Sn

= -0.14 V

Under standard conditions the potential for the reaction

Sn(s) + 2Fe3+ (aq) ® 2Fe 2+ (aq) + Sn 2+ (aq) is

EBD_7764 C-78

15.

(a) 0.91 V (b) 1.40 V [2004] (c) 1.68 V (d) 0.63 V The standard e.m.f. of a cell involving one electron change is found to be 0.591 V at 25ºC. The equilibrium constant of the reaction is (F = 96,500 C mol–1; R = 8.314 JK–1 mol–1) (a) 1.0 × 1010 (b) 1.0 × 105 [2004] 1 (c) 1.0 × 10 (d) 1.0 × 1030 The limiting molar conductivities Lº for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol–1 respectively. The Lº for NaBr is [2004] (a) 278 S cm2 mol–1 (b) 176 S cm2 mol–1 (c) 128 S cm2 mol–1 (d) 302 S cm2 mol–1 In a cell that utilises the reaction Zn(s) + 2H + (aq) ® Zn 2+ (aq) + H 2 (g) addition of H2SO4 to cathode compartment, will [2004] (a) increase the E and shift equilibrium to the right (b) lower the E and shift equilibrium to the right (c) lower the E and shift equlibrium to the left (d) increase the E and shift equilibrium to the left The E º 3+ 2 + values for Cr, Mn, Fe and Co

16.

are – 0.41, + 1.57, + 0.77 and + 1.97V respectively. For which one of these metals the change in oxidation state from +2 to +3 is easiest? (a) Fe (b) Mn [2004] (c) Cr (d) Co For a spontaneous reaction the D G, equilibrium

12.

13.

14.

17.

18.

M

/M

constant (K) and E oCell will be respectively [2005] (a) –ve, >1, –ve (b) –ve, 1, –ve (d) –ve, >1, +ve The highest electrical conductivity of the following aqueous solutions is of [2005] (a) 0.1 M difluoroacetic acid (b) 0.1 M fluoroacetic acid (c) 0.1 M chloroacetic acid (d) 0.1 M acetic acid Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is– Al 3+ + 3e- ® Al °

Chemistry To prepare 5.12 kg of aluminium metal by this method we require [2005]

(a) 5.49 × 101 C of electricity (b) 5.49 × 10 4 C of electricity (c) 1.83 × 10 7 C of electricity (d) 5.49 × 10 7 C of electricity 19.

KCl KNO 3 HCl NaOAc NaCl

Electrolyte: 2

–1

L ¥ (S cm mol ) : 149.9

145

426.2

91

126.5

¥

Calculate L HOAc using appropriate molar conductances of the electrolytes listed above at infinite dilution in H 2 O at 25°C (a) 217.5 (c) 552.7

[2005]

(b) 390.7 (d) 517.2 o

o

20. The molar conductivities L NaOAc and L HCl at infinite dilution in water at 25ºC are 91.0 and 426.2 S cm2/mol respectively. To calculate o

L HOAc , the additional value required is [2006] o

o

(b) L NaCl

o

(d) L KCl

(a)

L NaOH

(c)

LH O 2

o

21. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 W. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 W. The molar conductivity of 0.2 M solution of electrolyte will be [2006] (a) 1.24 × 10–4 S m2 mol–1 (b) 12.4 × 10–4 S m2 mol–1 (c) 124 × 10–4 S m2 mol–1 (d) 1240 × 10–4 S m2 mol–1 22. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below : [2007] LoCH3COONa = 91.0 S cm 2 / equiv. L o HCl = 426.2 S cm 2 / equiv.

What additional information/ quantity one needs

Electrochemistry

to calculate Lo of an aqueous solution of acetic acid? (a)

Lo of chloroacetic acid (ClCH2COOH)

(b)

Lo of NaCl

(c) L o of CH3COOK (d) the limiting equivalent coductance of H + (l°

H+

23.

).

The cell, Zn | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (E°cell = 1.10 V)

was allowed to be completely discharged at 298 K. The relative concentration of Zn 2+ to Cu2+ æ [Zn 2+ ] ö ç ÷ ç [Cu 2+ ] ÷ is è ø

[2007]

(a) 9.65 × 104 (c) 37.3 24.

25.

Given Eº

Cr 3+ / Cr

(b) antilog (24.08) (d) 1037.3. = –0.72 V, Eº

Fe2 + / Fe = – 0.42 V. The potential

for the cell Cr|Cr3+ (0.1M)|| Fe2 + (0.01 M)| Fe is [2008] (a) 0.26 V (b) 0.336 V (c) – 0.339 (d) 0.26 V In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH 3OH(l ) + 3/2O 2 (g) ¾¾ ®

CO2 (g) + 2H 2O(l )

26.

At 298 K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and and CO2 (g) are –166.2 –237.2 and –394.4 kJ mol–1 respectively. If standard enthalpy of combustion of methonal is – 726 kJ mol–1, efficiency of the fuel cell will be: [2009] (a) 87% (b) 90% (c) 97% (d) 80% Given: E° 3+ = –0.036V, Fe

E°

/ Fe

Fe 2+ / Fe

= –0.439 V

The value of standard electrode potential for the change, Fe3+ (aq) + e – ¾¾ ® Fe 2+ (aq) will be: [2009]

C-79 (a) 0.385 V (b) 0.770 V (c) –0.270 V (d) –0.072 V 27. The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows :

2 4 Al2 O3 ® Al + O2 , D r G = + 966 kJ mol -1 3 3 The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least [2010] (a) 4.5 V (b) 3.0 V (c) 2.5 V (d) 5.0 V 28. The correct order of E ° 2 + values with M

/M

negative sign for the four successive elements Cr, Mn, Fe and Co is [2010] (a) Mn > Cr > Fe > Co (b) Cr < Fe > Mn > Co (c) Fe > Mn > Cr > Co (d) Cr > Mn > Fe > Co 29. Resistance of 0.2 M solution of an electrolyte is 50 W. The specific conductance of the solution is 1.3 S m–1. If resistance of the 0.4 M solution of the same electrolyte is 260 W, its molar conductivity is : [2011RS] –4 2 –1 (a) 6.25 × 10 S m mol (b) 625 × 10–4 S m2 mol–1 (c) 62.5 S m2 mol–1 (d) 6250 S m2 mol–1 30. The standard reduction potentials for Zn2+/Zn, Ni2+/Ni and Fe2+/Fe are –0.76,–0.23 and –0.44 V respectively. The reaction X +Y 2 + ¾¾ ® X 2+ + Y will be spontaneous when : [2012] (a) X = Ni, Y = Fe (b) X = Ni, Y = Zn (c) X= Fe, Y = Zn (d) X= Zn, Y = Ni 31. Given : Eo 3+ = -0.74 V; Eo = 1.51 V Cr / Cr MnO 4 / Mn 2+ = 1.36 V Eo 2 - 3+ = 1.33 V; Eo Cr O / Cr Cl / Cl2 7

Based on the data given above, strongest oxidising agent will be : [2013] (a) Cl (b) Cr3+ (c) Mn 2+ (d) MnO4 – 32. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest Eo 3+ 2+ value ? M /M

[2013]

EBD_7764 C-80

33.

34.

Chemistry

(a) Cr(Z = 24) (b) Mn(Z = 25) (c) Fe(Z = 26) (d) Co(Z = 27) Resistance of 0.2 M solution of an electrolyte is 50 W. The specific conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of the same electrolyte is 280 W. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol–1 is: [2014] (a) 5 × 10–4 (b) 5 × 10–3 (c) 5 × 103 (d) 5 × 102 The equivalent conductance of NaCl at concentration C and at infinite dilution are lC and l¥ , respectively. The correct relationship between lC and l¥ is given as: (Where the constant B is positive)

35.

[2014]

(

)

2 Mn3+ + e - ® Mn 2+ ; E o = +1.51V The Eo for 3Mn 2 + ® Mn + 2Mn 3+ will be: [2014] (a) –2.69 V; the reaction will not occur (b) –2.69 V; the reaction will occur (c) –0.33 V; the reaction will not occur (d) –0.33 V; the reaction will occur 36. Two Faraday of electricity is passed through a solution of CuSO 4 . The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu) [JEE M 2015] (a) 2g (b) 127 g (c) 0 g (d) 63.5 g 37. Given [JEE M 2017]

(a)

lC = l ¥ + ( B ) C

Eo

(b)

lC = l ¥ - ( B ) C

o

(c)

lC = l¥ - ( B ) C

(d)

lC = l¥ + ( B ) C

Cl /Cl –

2

E

= 1.36V, E o 3 + = –0.74V, Cr / Cr

Cr /O2– /Cr3+ 2 7

= 1.33V, Eo

– /Mn 2 + MnO4

= 1.51V.

Among the following, the strongest reducing agent is (a) Cr (b) Mn 2+ (c) Cr3+ (d) Cl–

Given below are the half-cell reactions: Mn 2 + + 2e - ® Mn ; E o = -1.18V

Answer Key 1 (b) 16 (d) 31 (d)

1.

2.

2 (c) 17 (a) 32 (d)

3 (b) 18 (d) 33 (a)

(b) given S µ

4 (a) 19 (b) 34 (c)

5 (d) 20 (b) 35 (a)

6 (c) 21 (b) 36 (d)

7 (d) 22 (b) 37 (a)

8 (a) 23 (d)

area ´ conc km 2 mol = l m ´ m3

\k = Sm2 mol-1 (c) Ecell = Reduction potential of cathode (right) – reduction potential of anode (left) = Eright – Eleft.

9 (b) 24 (d)

3.

10 (b) 25 (c)

11 (a) 26 (b)

12 (a) 27 (c)

13 (c) 28 (a)

14 (a) 29 (a)

(b) Oxidation half call:H2(g) ––––––– ® – 2H+(1M) + 2e– P1 Reduction half cell 2H+ (1M) + 2e– ––––––– ® – H2(g) P2 The net cell reaction H2(g) ––––––– ® – H2(g) P2 P1

15 (c) 30 (d)

C-81

Electrochemistry

E ºcell = 0.00 V

n=2

11.

(a)

RT logeK nF P RT log e 2 =0– P1 nF

\ Ecell = Eºcell –

or Ecell = 4.

5. 6.

for Sn(s) + 2Fe3+ (aq) ® 2Fe 2 + (aq) + Sn 2 + (aq) \ Standar d potential for the given reaction

(a)

log K = Eº cell ´ log K =

7.

Sn 2+ + 2e - ® Sn(s) DG° = -2 ´ F(-0.14)

P2 RT log e P 2F 1

2Cr 3+ + 7 H 2 O ® Cr2 O 72 - + 14H + O.S. of Cr changes from +3 to +6 by loss of electrons. At anode oxidation takes place. (d) Pure metal always deposits at cathode. (c) The equilibrium constant is related to the standard emf of cell by the expression

(d)

A +0.5C

or E ocell = E o

Sn / Sn 2 +

12. (a)

590 = 10 or K = 1 × 1010 59

C –1.2V

8.

= 9.

(b)

13. (c)

0.059 [Cu +2 ] log n [ Zn + 2 ]

0.059 log[0.1] 2 = 1.10 - 0.0295 = 1.07 V = 1.10 +

10.

(b) In H 2 - O 2 fuel cell, the combustion of H2 occurs to create potential difference between the two electrodes

0.059 log K c n

or 0 = 0.591 –

0.0591 log K c 1

0.591 = 10 or Kc = 1 ´ 1010 0.0591

L º NaCl = l° Na + + lCl -

....(i)

L° KBr = l°K + + l°Br -

....(ii)

L° KCl = l°K + + lCl-

....(iii)

operating (i) + (ii) - (iii) L° NaBr = l° Na + + l º Br –

= 126 + 152 - 150 = 128 S cm 2 mol -1

14. (a)

108 ´ 9650 = 10.8 g 96500

E cell = E º cell +

Fe3+ / Fe 2 +

E°cell = Eocell –

or log K c =

NOTE The higher the negative value of

reduction potential, the more is the reducing power. Hence B > C > A. (a) When 96500 coulomb of electricity is passed through the electroplating bath the amount of Ag deposited = 108g \ when 9650 coulomb of electricity is passed deposited Ag.

+ Eo

= 0.14 + 0.77 = 0.91 V

n 2 = 0.295 ´ 0.059 0.059

B –3.0V

Fe3+ + e - ® Fe2+ DG° = -1´ F ´ 0.77

Zn(s) + 2H + + (aq)

E cell = E °cell -

Zn 2+ (aq) + H 2 (g)

[Zn 2 + ][H 2 ] 0.059 log 2 [H + ]2

Addition of H2SO4 will increase [H+]and Ecell will also increase and the equilibrium will shift towards RHS 15. (c) The given values show that Cr has maximum oxidation potental, therefore its oxidation will be easiest. (Change the sign to get the oxidation values) 16. (d)

NOTE For spontaneous reaction DG

should be negative. Equilibrium constant should be more than one

EBD_7764 C-82

17.

18.

(DG = – 2.303 RT log Kc, If Kc = 1 then DG = 0; If Kc < 1

Chemistry m (molar conductivity) = ? m = k × V (k can be calculated as

then DG = +ve). Again DG = - nFE ocell .

k=

E ocell must be +ve to have DG –ve.

cell constant is known.) Hence,

(a) Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid > fluoro acetic acid > chloro acetic acid > acetic acid. (d) 1 mole of e– = 1F = 96500 C 27g of Al is deposited by 3 × 96500 C 5120 g of Al will be deposited by =

q= \Q

19.

(b)

5120 ´ 96500 ´ 3 = 5.49 ´ 107 C 27

L¥ HCl = 426.2

(i)

L¥ AcONa = 91.0

(ii)

L¥ NaCl = 126.5

(iii)

equation

(

o = L CH

3COONa

)(

+ L oHCl - L oNaCl

o Hence L NaCl is required.

21.

(b)

1ælö R = 100 W , k = ç ÷ R è aø

22. (b)

NOTE According to Kohlrausch’s law,,

molar conductivity of weak electrolyte acetic acid (CH3COOH) can be calculated as follows:

(

)

L oCH3COOH = L oCH3COONa + L oHCl - L oNaCl

Ecell = E°cell

,

l (cell constant) = 1.29 × 100m–1 a

Given, R = 520W, C = 0.2 M,

æ é Zn 2 + ù ö 0.059 û÷ log ç ë ç ÷ + 2 2 ç é Cu ù ÷ ûø èë

æ é Zn 2 + ù ö 0.059 û÷ log ç ë or 0 = 1.1 ç é 2+ ù ÷ 2 ç Cu ÷ ûø èë æ é Zn 2+ ù ö û ÷ = 2 ´1.1 = 37.3 log ç ë ç é 2+ ù ÷ 0.059 ç Cu ÷ ûø èë

o L CH is given by the following 3COOH

LoCH COOH 3

= 12.4 ×10–4 Sm2 mol–1

23. (d) E cell = 0; when cell is completely discharged.

mFz M

= [426.2 + 91.0 - 126.5] = 390.7 (b)

1 1000 ´ 129 ´ ´ 10 -6 m 3 520 0.2

for calculating value of LoCH 3COOH .

¥ L AcOH = (i) + (ii) - (iii)

20.

m=

now

\ Value of L o NaCl should also be known

3 ´ 96500 ´ 5120 = 5.49 ´ 10 7 C 27

q= We know, Q

1 æ1ö R çè a ÷ø

)

æ é Zn 2 + ù ö û ÷ = 1037.3 çë \ ç 2+ ù ÷ é ç Cu ÷ ûø èë

24. (d) From the given representation of the cell, Ecell can be found as follows. 0.059 [Cr3+ ] log 6 [Fe2+ ]3

2

Ecell = Eo

2+

Fe

/ Fe

- Eo

Cr

3+

/ Cr

-

[Nernst -Equ.] = –0.42 – (–0.72) -

0.059 (0.1)2 log 6 (0.01)3

C-83 \ The potential difference needed for the reduction = 2.5 V.

Electrochemistry

= –0.42 + 0.72 -

0.059 0.1´ 0.1 log 6 0.01´ 0.01´ 0.01

0.059 10-2 0.059 log = 0.3 ´4 -6 6 6 10

= 0.3 -

28. (a) The value of E o

M 2∗ M

are

= 0.30 – 0.0393 = 0.26 V Hence option (d) is correct answer. 25.

(c) CH3OH (l) +

Eo

Mn 2∗ Mn

3 O (g) ® CO2 (g) + 2H2O 2 2

< ,0.44 V and

Co 2∗ Co

M

Fe

3+

29. (a)

/ Fe

… (i)

= –0.036 V

Fe 2 + + 2e - ® Fe ,

E°

Fe 2+ / Fe

= –0.439V

… (ii)

we have to calculate Fe3+ + e - ® Fe2 + , DG = ?

To obtain this equation subtract equ (ii) from (i) we get Fe3+ + e – ® Fe 2 + As we know that DG = –nFE Thus for reaction (iii)

… (iii)

DG = DG1 - DG – nFE° = – nFE1 – (–nFE2) –nFE° = nFE2 – nFE1 –1FE° = 2× 0.439F – 3 × 0.036 F –1 FE° = 0.770 F \ E° = – 0.770V O -- > F- > Na + > Mg ++ > Al3+

27.

(c) DG = – nFE or E
Cr2+ > Fe2+ > Co2+. k=

Fe3+ + 3e - ® Fe ,

E°

< ,0.28 V.

The correct order of E o 2∗

702.6 ´ 100 = 97% 726

(b) Given

Eo Eo

3 DG f (CH3OH, l ) – DG f (O2 ,g) 2 = – 394.4 + 2 (–237.2) – (–166.2) – 0 = – 394.4 – 474.4 + 166.2 = – 702.6 k J

26.

< ,0.9 V,

Fe 2∗ Fe

DG r = DG f (CO 2 , g) + 2DG f (H 2 O, l) -

< ,1.18 V,

Eo

Cr 2∗ Cr

(l)

% efficiency =

for given metal ions

1 l ´ R A

1.3 =

1 l ´ 50 A

l = 65m-1 A k ´ 1000 L= molarity [molarity is in moles/litre but 1000 is used to convert liter into cm3]

æ 1 ö ´ 65 m -1 ÷ ´ 1000 cm 3 çè ø 260 = 0.4 moles

650 m -1 1 ´ m3 260 ´ 4 mol 1000 = 6.25 × 10–4 S m2 mol–1 30. (d) For a spontaneous reaction DG must be –ve Since DG = – nFE° Hence for DG to be -ve DE° has to be positive. Which is possible when X = Zn, Y = Ni =

® Zn++ + Ni Zn + Ni++ ¾¾ E°

Zn / Zn +2

+ E°

Ni 2+ / Ni

EBD_7764 C-84

31.

32.

Chemistry

= 0.76 + (– 0.23) = + 0.53 (positive) (d) Higher the value of standard reduction potential, stronger is the oxidising agent, hence MnO4– is the strongest oxidising agent. (d)

E°

Cr 3+ / Cr 2 +

0.77 V E ° 3+ Mn

E°

33.

= – 0.41 V E°Fe3+ / Fe 2 + = +

/ Mn 2 +

Co3+ / Co 2 +

= + 1.57 V,,

2.5 ´ 10-3 ´ 1000 0.5 = 5 S cm2 mol–1 = 5 × 10–4 S m2 mol–1 34. (c) According to Debye Huckle onsager equation, =

lC = l¥ - B C 35. (a) (a)

Mn 2+ + 2e- ® Mn; E o = -1.18V ;

= + 1.97 V

(b) Mn 3+ + e ® Mn 2+ ; E o = -1.51V ; ...(ii)

(a) Given for 0.2 M solution R = 50 W k = 1.4 S m–1 = 1.4 × 10–2 S cm–1 Now, R = r

Þ Þ

1 l l R=r = ´ a k a

k=

Now multiplying equation (ii) by two and subtracting from equation (i)

l 1 l = ´ a k a

l -2 Þ = R ´ k = 50 ´1.4 ´ 10 a For 0.5 M solution R = 280 W k =? l = 50 ´ 1.4 ´ 10-2 a

1 ´ 50 ´ 1.4 ´ 10 -2 280

1 = ´ 70 ´ 10-2 280 = 2.5 × 10–3 S cm–1

k ´1000 Now, L m = M

...(i)

3Mn 2+ ® Mn + + 2Mn 3+ ; E o = EOx. + ERed.

= – 1.18 + (– 1.51) = – 2.69 V (–ve value of EMF (i.e. DG = +ve) shows that the reaction is non-spontaneous) 36.

(d)

® Cu Cu 2+ + 2e – ¾¾ 2F i.e. 2 × 96500 C deposit Cu = 1 mol = 63.5g

37. (a)

E° E° E°

E°

– / Mn 2 + MnO 4 Cl2 /Cl –

= 1.36V

Cr2 O72 - /Cr3+

Cr3+ / Cr

= 1.51V

= 1.33V

= – 0.74

Since Cr 3+ is having least reducing potential, so Cr is the best reducing agent.

C-85

Chemical Kinetics

18

Chemical Kinetics 1.

2.

3.

Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively [2002] (a) sec–1 , Msec–1 (b) sec–1, M (c) Msec–1 , sec–1 (d) M, sec–1. For the reaction A + 2B ® C, rate is given by R = [A] [B]2 then the order of the reaction is [2002] (a) 3 (b) 6 (c) 5 (d) 7. The differential rate law for the reaction H2 + I2 ® 2HI is [2002] -

(b)

d[H 2 ] d[I 2 ] 1 d[Hl ] = = dt dt 2 dt

(c)

1 d[H 2 ] 1 d[I 2 ] d[Hl] = =2 dt 2 dt dt

d[H 2 ] d[I 2 ] d[HI] = -2 = dt dt dt If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is [2002] (a) 16 grams (b) 2 grams (c) 32 grams (d) 8 grams. The integrated rate equation is [2002] Rt = log C0 - logCt. The straight line graph is obtained by plotting

(d)

4.

5.

d[ H 2 ] d[ I 2 ] d[ HI ] ==dt dt dt

(a)

-2

(a) time vs logCt

(b)

1 vs C t time

(c) time vs Ct

(d)

1 1 vs time Ct

6.

The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of it remaining undecayed after 18 hours would be [2003] (a) 8.0 g (b) 12.0 g (c) 16.0 g (d) 4.0 g 7. In respect of the equation k = Ae - E a / RT in chemical kinetics, which one of the following statements is correct ? [2003] (a) A is adsorption factor (b) Ea is energy of activation (c) R is Rydberg’s constant (d) k is equilibrium constant 8. For the reaction system : [2003] 2NO(g) + O 2 (g) ® 2 NO 2 (g) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will (a) diminish to one-eighth of its initial value (b) increase to eight times of its initial value (c) increase to four times of its initial value (d) diminish to one-fourth of its initial value 9. In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M is 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is [2004] (a) 7.5 minutes (b) 15 minutes (c) 30 minutes (d) 60 minutes 10. The rate equation for the reaction 2A + B ® C is found to be: rate = k[A][B]. The correct statement in relation to this reaction is that the [2004] (a) rate of formation of C is twice the rate of disappearance of A (b) t1 / 2 is a constant (c) unit of k must be s–1 (d) value of k is independent of the initial concentrations of A and B

EBD_7764 C-86 11. The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is [2004] (a) 3.125 g (b) 2.084 g (c) 1.042 g (d) 4.167 g 12. A reaction involving two different reactants can never be [2005] (a) bimolecular reaction (b) second order reaction (c) first order reaction (d) unimolecular reaction 13. A schematic plot of ln K eq versus inverse of

temperature for a reaction is shown below [2005]

ln Keq

6.0

2.0 1.5 ´ 10 - 3 1 (K -1 ) 2.0 ´ 10 - 3 T

The reaction must be (a) highly spontaneous at ordinary temperature (b) one with negligible enthalpy change (c) endothermic (d) exothermic 14.

t 1 can be taken as the time taken for the 4

3 concentration of a reactant to drop to of its 4 initial value. If the rate constant for a first order

reaction is K, the

15.

t1 4

can be written as

(a) 0.75/K (b) 0.69/K [2005] (c) 0.29/K (d) 0.10/K Consider an endothermic reaction X ® Y with the activation energies E b and E f for the backward and forward reactions, respectively. In general [2005] (a) there is no definite relation between E b and E f (b)

E b = Ef

(c)

E b > Ef

(d)

Eb < Ef

Chemistry 16. A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will [2006] (a) increase by a factor of 4 (b) double (c) remain unchanged (d) triple 17. Rate of a reaction can be expressed by Arrhenius equation as : [2006] k = A e–E/RT In this equation, E represents (a) the total energy of the reacting molecules at a temperature, T (b) the fraction of molecules with energy greater than the activation energy of the reaction (c) the energy above which all the colliding molecules will react (d) the energy below which all the colliding molecules will react 18. The following mechanism has been proposed for the reaction of NO with Br 2 to form NOBr :

NO(g) + Br2(g)

NOBr2(g)

NOBr 2 (g ) + NO (g ) ¾ ¾® 2 NOBr ( g )

If the second step is the rate determining step, the order of the reaction with respect to NO(g) is [2006] (a) 3 (b) 2 (c) 1 (d) 0 19. The energies of activation for forward and reverse reactions for A2 + B2 ƒ 2AB are 180 kJ mol–1 and 200 kJ mol–1 respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change of the reaction (A2 + B2 ® 2AB) in the presence of a catalyst will be (in kJ mol–1) [2007] (a) 20 (b) 300 (c) 120 (d) 280 20. Consider the reaction, 2A + B ® products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate

Chemical Kinetics

21.

22.

increased by two times. The unit of rate constant for this reaction is [2007] (a) s – 1 (b) L mol–1 s–1 (c) no unit (d) mol L–1 s–1. A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial velocity is ten times the permissible value, after how many days will it be safe to enter the room? [2007] (a) 100 days (b) 1000 days (c) 300 days (d) 10 days. 1 A ® 2B, rate of disappearance 2 of ‘A’ is related to the rate of appearance of ‘B’ by the expression [2008]

For a reaction

(a)

24.

25.

d[A] 1 d[B] d[A] 1 d[B] (b) – = = dt 2 dt dt 4 dt

d[A ] d[B] d[A] d[B] (d) – = =4 dt dt dt dt The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301) [2009] (a) 23.03 minutes (b) 46.06 minutes (c) 460.6 minutes (d) 230.03 minutes The time for half life period of a certain reaction ® Products is 1 hour. When the initial A ¾¾ concentration of the reactant ‘A’, is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction ? [2010] (a) 4 h (b) 0.5 h (c) 0.25 h (d) 1 h Consider the reaction : Cl 2 (aq) + H 2S(aq) ® S(s) + 2H + (aq) + 2Cl - (aq) The rate equation for this reaction is rate = k[Cl 2 ][H 2S] Which of these mechanisms is/are consistent with this rate equation? [2010] + + A. Cl 2 + H 2S ® H + Cl + Cl + HS (slow)

(c)

23.

– –

Cl + + HS- ® H + + Cl- + S

B.

(fast)

H 2S ƒ H + + HS- (fast equilibrium)

Cl 2 + HS- ® 2Cl - + H + + S (Slow) (a) B only (b) Both A and B (c) Neither A nor B (d) A only

C-87 26. A reactant (A) froms two products : [2011RS] k

1 ® B, Activation Energy Ea A ¾¾ 1

k

2 A ¾¾® C, Activation Energy Ea2

If Ea2 = 2 Ea1, then k1 and k2 are related as : (a)

k2 = k1e Ea1 / RT

(b) k2 = k1e Ea2 / RT

(c) k1 = Ak2e Ea1 / RT (d) k1 = 2k2e Ea2 / RT 27. For a first order reaction (A) ® products the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is : [2012] (a) 1.73 × 10–5 M/min (b) 3.47 × 10–4 M/min (c) 3.47 × 10–5 M/min (d) 1.73 × 10–4 M/min 28. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be : (R = 8.314 JK–1 mol–1 and log 2 = 0.301) [2013] (a) 53.6 kJ mol –1 (b) 48.6 kJ mol–1 –1 (c) 58.5 kJ mol (d) 60.5 kJ mol–1 29. For the non - stoichiometric reaction 2A + B ® C + D, the following kinetic data were obtained in three separate experiments, all at 298 K. Initial Con centration (A )

Initial Concentration (B )

0.1 M

0.1 M

In itial rate of formation of C –1 –1 (mol L s ) 1.2 × 10

–3

0.1 M

0.2 M

1.2 × 10

–3

0.2 M

0.1 M

2.4 × 10

–3

The rate law for the formation of C is: (a)

(b)

dc 2 = k [ A] [ B ] dt

dc dc 2 = k [ A][ B ] = k [ A] (d) dt dt Higher order (>3) reactions are rare due to : [JEE M 2015] (a) shifting of equilibrium towards reactants due to elastic collisions (b) loss of active species on collision (c) low probability of simultaneous collision of all the reacting species (d) increase in entropy and activation energy as more molecules are involved

(c)

30.

dc = k [ A][ B ] dt

[2014]

EBD_7764 C-88 31. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be: [JEE M 2016] –1 (a) 2.66 L min at STP

Chemistry

32. Two reactions R1 and R2 have identical preexponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R = 8.314 J mol–1K–1)

[JEE M 2017]

(b) 1.34 × 10–2 mol min–1

(a) 8

(b) 12

(c)

6.96 × 10–2 mol min–1

(c) 6

(d) 4

(d)

6.93 × 10–4 mol min–1

Answer Key 1 (a) 16 (a) 31 (d)

1.

2.

2 (a) 17 (c) 32 (d)

3 (d) 18 (b)

4 (d) 19 (a)

5 (a) 20 (b)

6 (d) 21 (a)

7 (b) 22 (b)

(a) For a zero order reaction. rate =k[A]º i.e. rate = k hence unit of k = M.sec–1 For a first order reaction. rate = k [A] k = M.sec–1/M = sec–1 (a)

4.

NOTE Order is the sum of the power of

(d) rate of appearance of HI = rate of formation of H2 = rate of formation of I2 = hence

-d [ H 2 ] dt

=-

dt

5.

1 d [ HI] 2 dt

-d éë H 2 ùû dt

-d [ I 2 ]

d [ I2 ]

9 (c) 24 (c)

10 (d) 25 (d)

or –

the concentrations terms in rate law expression. Hence the order of reaction is = 1 + 2 = 3 3.

8 (b) 23 (b)

dt =

1 d [ HI] 2 dt

11 (a) 26 (c)

2d [ H 2 ]

12 (d) 27 (b)

=-

13 (d) 28 (a)

2d [ I2 ]

14 (c) 29 (d)

=

d [ HI]

dt dt dt (d) t1/2 = 5 years, T = 15 years hence total number of half life periods = 15 = 3 . 5 \ Amount left = 64 = 8g ( 2) 3 (a) Rt = log Co – log Ct It is clear from the equation that if we plot a graph between log Ct and time, a straight

k and 2.303 intercept equal to log [Ao] will be obtained. (d) t1/2 = 3hrs. T = 18 hours Q T = n ´ t1/ 2

line with a slope equal to –

6.

15 (d) 30 (c)

18 =6 3 Initial mass (C0) = 256 g \n=

C-89

Chemical Kinetics

\ Cn =

7.

8.

9.

10.

11.

C0 2

n

=

256 ( 2)

6

=

256 = 4g. 64

(b) In equation k = Ae - E a / RT ; A = Frequency factor k = velocity constant, R = gas constant and Ea = energy of activation (b) r = k [O2][NO]2. When the volume is reduced to 1/2, the conc. will double \ New rate = k [2O2][2 NO]2 = 8 k [O2][NO]2 The new rate increases to eight times of its initial. (c) As the concentration of reactant decreases from 0.8 to 0.4 in 15 minutes hence the t1 / 2 is 15 minutes. To fall the concentration from 0.1 to 0.025 we need two half lives i.e., 30 minutes. (d) The velocity constant depends on temperature only. It is independent of concentration of reactants. æ 1ö (a) Nt = N 0 ç ÷ è 2ø

n

where n is number of half

life periods. n=

Total time 24 = =6 half life 4 6

æ 1ö \ N t = 200 ç ÷ = 3.125g . è 2ø 12.

14. (c)

=

2.303 (2 log 2 - log 3) K

(f)

(b)

(b)

(f)

16. (a) Since the reaction is 2nd order w.r.t CO. Thus, rate law is given as. r = k [CO]2 Let initial concentration of CO is a i.e. [CO] =a \ r1 = k (a)2 = ka2 when concentration becomes doubled, i.e.[CO] = 2a \ r2 = k (2a)2 = 4ka2 \ r2 = 4r1 So, the rate of reaction becomes 4 times. 17. (c) In Arrhenius equation k = A e–E/RT, E is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products. 18. (b) (i) NO(g) + Br2(g) NOBr2(g) ¾® 2 NOBr ( g ) (ii) NOBr2 (g ) + NO (g ) ¾ Rate law equation = k[NOBr2] [NO] But NOBr2 is intermediate and must not appear in the rate law equation

from Ist step K C =

[NOBr2 ] [NO] [Br2 ]

\ [NOBr2] = KC [NO] [Br2] \ Rate law equation = k . KC [NO]2 [Br2] hence order of reaction is 2 w.r.t. NO.

log k =

1 would be negative straight T line with positive slope.

2.303 (log 4 - log 3) K

for an endothermic reaction DH = +Ve hence for DH to be negative Ea < Ea

different reactant can never be unimolecular. (d) The graph show that reaction is exothermic.

\ log k Vs

=

2.303 0.29 (2 ´ 0.301 - 0.4771) = K K 15. (d) Enthalpy of reaction (DH) = Ea – Ea

(d) The molecularity of a reaction is the number of reactant molecules taking part in a single step of the reaction.

-DH +1 RT For exothermic reaction DH < 0

2.303 1 2.303 4 log = log K K 3/ 4 3

=

NOTE The reaction involving two

13.

t1/ 4 =

19. (a)

DH R = E f - E b = 180 – 200 = – 20 kJ/mol The nearest correct answer given in choices may be obtained by neglecting sign.

EBD_7764 C-90

20.

21.

Chemistry

(b) For a first order reaction t1/ 2 =

0.693 i.e. K

for a first order reaction t1/ 2 does not depend up on the concentration. From the given data, we can say that order of reaction with respect to B = 1 because change in concentration of B does not change half life. Order of reaction with respect to A = 1 because rate of reaction doubles when concentration of B is doubled keeping concentration of A constant. \ Order of reaction = 1 + 1 = 2 and units of second order reaction are L mol–1 sec–1. (a) Suppose activity of safe working = A Given A0 = 10A l=

0.693 0.693 = t1/ 2 30

t½ =

A 2.303 2.303 10A log 0 = log l A 0.693 / 30 A

2.303 ´ 30 ´ log10 = 100 days. 0.693 (b) The rates of reactions for the reaction 1 A ¾¾ ® 2B 2 can be written either as

=

22.

0.693 2.303 ´ 2 = 6.93 t

t = 46.06 min 24. (c) For the reaction A ® Product given t1/ 2 = 1 hour for a zero order reaction

\ t1 / 2 = or k =

k=

25.

[ A0 ] 2 t1/ 2

=

2 = 1 mol lit –1 hr–1 2 ´1

dx change in concentration = dt time

Rate = k [ Cl2 ][ H 2S] Now if we consider option (2) we find Rate = k [ Cl2 ] éë HS ùû

...(1)

From equation (i) k=

é H + ù éHS- ù ë ûë û H 2S

or éë HS ùû =

k [ H 2 S] H+

Substituting this value in equation (1) we find

d 1 d [A] = [B] dt 4 dt i.e., correct answer is (b) (b) For first order reaction,

or –

0.693 2.303 100 log = 6.93 t 1

2k

0.50 - 0.25 time \ time = 0.25 hr. (d) Since the slow step is the rate determining step hence if we consider option (1) we find

d 1 d –2 [A] = [B] dt 2 dt

2.303 100 log t 100 - 99

[ A0 ]

1=

1 d [B] with respect to ‘B’ or 2 dt From the above, we have

k=

k

initial conc. rate constant

Further for a zero order reaction

d -2 [A] with respect to ‘A’’ dt

23.

[ A0 ] =

tcompletion =

Rate = k [ Cl2 ] K

[ H 2S] = k ' [ Cl2 ][ H 2S] H+

éH+ ù ë û hence only , mechanism (1) is consistent with the given rate equation.

26. (c)

k1 = A1e - Ea1 / RT

.........(i)

k2 = A2 e- Ea2 / RT

........(ii)

C-91

Chemical Kinetics

On dividing eqn (i) from eqn. (ii)

We find, x = 1

k1 A = 1 ( Ea1 - Ea1 ) / RT k 2 A2

Hence

........(iii)

30.

(c)

Reactions of higher order (>3) are very rare due to very less chances of many molecules to undergo effective collisions.

31.

(d)

H2O2(aq) ® H2O(aq) +

Given Ea2 = 2 Ea1 On substituting this value in eqn. (iii) k1 = k 2 A ´ e 27.

Ea / RT 1

2.0303 a 2.303 0.1 log k= = log t a-x 40 0.025

28.

2.303 2.303 ´ 0.6020 log 4 = 40 40

= 3.47 ´ 10 –2 R = k (A)1 = 3.47 × 10–2 × 0.01 = 3.47 × 10–4 (a) Activation energy can be calculated from the equation

log k2 - Ea æ 1 1 ö = log k1 2.303 R çè T2 T1 ÷ø

Given

k2 = 2 ; T2 = 310 K ; T1 = 300 K k1

- Ea 1 ö æ 1 ç ÷ 2.303 ´ 8.314 è 310 300 ø Ea = 53598.6 J/mol = 53.6 kJ/mol.

d [C ] = k[A]x [B]y t Now from the given data 1.2 × 10 – 3 = k [0.1]x[0.1]y .....(i) 1.2 × 10 – 3 = k [0.1]x[0.2]y .....(ii) 2.4 × 10 – 3 = k [0.2]x[0.1]y .....(iii) Dividing equation (i) by (ii)

(d) Let rate of reaction =

Þ

-3

1.2 ´10 1.2 ´10

-3

=

x

k[0.1] [0.1]

Þ

2.4 ´10 -3

=

2.303 a log t (a - x)

Given a = 0.5, (a – x) = 0.125, t = 50 min \

k=

2.303 0.5 log 50 0.125

= 2.78 ´ 10–2 min–1 r = k[H2O2] = 2.78 ´ 10–2 ´ 0.05 = 1.386 ´ 10–3 mol min–1 Now

-

\

d [ H 2 O2 ]

=

dt 2d [ O2 ]

= -

dt d [ O2 ] dt

d [ H2 O]

=

dt

=

2d [ O2 ] dt

d [ H 2 O2 ] dt

1 d[H 2 O 2 ] ´ 2 dt

1.386 ´10-3 = 6.93 ´ 10–4 mol min–1 2 32. (d) From arrhenius equation, =

– Ea

k = A.e RT so, k1 = A.e

–E a /RT 1

k 2 = A.e

–E a / RT 2

.....(1) .....(2)

On dividing equation (2) (1)

y

k [0.1]x [0.2] y

We find, y = 0 Now dividing equation (i) by (iii) 1.2 ´10-3

k=

\

= log 2 =

29.

1 O (g) 2 2

For a first order reaction

(b) For a first order reaction

=

d [C ] = k[ A]1 [ B ]0 dt

k[0.1]x [0.1] y k [0.2] x [0.1] y

k Þ 2 =e k1

(E a – Ea ) 1 2 RT

10, 000 æ k ö Ea – Ea 2 ln ç 2 ÷ = 1 = =4 RT 8.314 ´ 300 è k1 ø

EBD_7764 C-92

Chemistry

19

Surface Chemistry 1.

2.

3.

4.

The formation of gas at the surface of tungsten due to adsorption is the reaction of order [2002] (a) 0 (b) 1 (c) 2 (d) insufficient data. Which one of the following characteristics is not correct for physical adsorption ? [2003] (a) Adsorption increases with increase in temperature (b) Adsorption is spontaneous (c) Both enthalpy and entropy of adsorption are negative (d) Adsorption on solids is reversible Identify the correct statement regarding enzymes [2004] (a) Enzymes are specific biological catalysts that cannot be poisoned (b) Enzymes are normally heterogeneous catalysts that are very specific in their action (c) Enzymes are specific biological catalysts that can normally function at very high temperatures (T~1000K) (d) Enzymes are specific biological catalysts that possess well-defined active sites

7.

8.

The volume of a colloidal particle, VC as compared to the volume of a solute particle in a true solution VS , could be VC VS VC VS

[2005]

VC ~ - 3 - 10 VS VC ~ ~ 10 23 (c) (d) -1 VS The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statements is NOT correct ? [2005] (a) Coagulation in both sols can be brought about by electrophoresis

(a)

5.

6.

~ 10 3 -

(b)

9.

(b) Mixing the sols has no effect (c) Sodium sulphate solution causes coagulation in both sols (d) Magnesium chloride solution coagulates, the gold sol more readily than the iron (III) hydroxide sol In Langmuir's model of adsorption of a gas on a solid surface [2006] (a) the mass of gas striking a given area of surface is proportional to the pressure of the gas (b) the mass of gas striking a given area of surface is independent of the pressure of the gas (c) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered (d) the adsorption at a single site on the surface may involve multiple molecules at the same time Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 amd 0.005, respectively. the correct order of their protective powers is [2008] (a) D < A < C < B (b) C < B < D < A (c) A < C < B < D (d) B < D < A < C Which of the following statements is incorrect regarding physissorptions? [2009] (a) More easily liquefiable gases are adsorbed readily. (b) Under high pressure it results into multi molecular layer on adsorbent surface. (c) Enthalpy of adsorption ( DH adsorption) is low and positive. (d) It occurs because of van der Waal’s forces. According to Freundlich adsorption isotherm which of the following is correct? [2012] x µ p0 (a) m x µ p1 (b) m

Surface Chemistry

x µ p1/n m (d) All the above are correct for different ranges of pressure The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order : [2013] (a) Al3+ < Ba2+ < Na+ (b) Na+ < Ba2+ < Al3+ (c) Ba2+ < Na+ < Al3+ (d) Al3+ < Na+ < Ba2+ For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants) [JEE M 2016] (a) Only 1 n appears as the slope.

(c)

10.

11.

C-93 12. The Tyndall effect is observed only when following conditions are satisfied: [JEE M 2017]

(i)

The diameter of the dispersed particles is much smaller than the wavelength of the light used.

(ii) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (iii) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

(b) log (1 n ) appears as the intercept.

(iv) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(c) Both k and 1 n appear in the slope term.

(a) (i) and (iv)

(b) (ii) and (iv)

(d) 1 n appears as the intercept.

(c) (i) and (iii)

(d) (ii) and (iii)

Answer Key 1 (b)

1.

2 (a)

3.

4.

4 (a)

5 (b)

6 (a)

7 (c)

(b) It is zero order reaction [

2.

3 (d)

8 (c)

5.

NOTE Adsorption of gas on metal sur--

face is of zero order] (a) As adsorption is an exothermic process. \ Rise in temperature will decrease adsorption (according to Le-chatelier principle). (d) Enzymes are very specific biological catalysts possessing well - defined active sites (a) Particle size of colloidal particle = 1mµ to 100 mµ (suppose 10 mµ) Vc =

4 3

pr

3

4 3 Vc = p(10) = 3

Particle size of true solution particle = 1mm Vs =

4 3

p(1)

3

hence now

Vc Vs

= 10

3

6.

9 (d)

10 (c)

11 (a)

12 (b)

(b) When oppositely charged sols are mixed their charges are neutralised. Both sols may be partially or completely precipitated. (a) According to Langmuir's Model of adsorption of a gas on a soild surface the mass of gas adsorbed(x)per gram of the adsorbent (m) is directly proportional to the pressure of the gas (p) at constant temperature i.e.

x

7.

µp m (c) For a protective colloid µ lesser the value of gold number better is the protective power. Thus the correct order of protective power of A, B, C and D is Þ (A) < (C) < (B) < (D) Gold number 0.50 0.10 0.01 0.005 Hence (c) is the correct answer

EBD_7764 C-94 8. (c) Adsorption is an exothermic process, hence DH will always be negative 9. (d) The Freundlich adsorption isotherm is mathematically represented as

Chemistry 11.

(a)

log

log x/m

at high pressure 1/n = 0. Hence, x / m µ P° at low pressure 1/n = 1 Hence, x/m µ P¢ (c) According to Hardy Schulze rule, greater the charge on cation, greater is its coagulating power for negatively charged sol (As2S3), hence the correct order of coagulating power : Na+ < Ba2+ < Al3+

x 1 = log K + log P m n

Thus if a graph is plotted between log(x/m) and log P, a straight line will be obtained

x = kP1/n m

10.

According to Freundlich adsorption isotherm

s lo

= pe

1/

n

Intercept = log K log P

The slope of the line is equal to 1/n and the intercept on log x/m axis will correspond to log K.

12. (b)

C-95

General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements 1.

2.

3.

4.

5.

Aluminium is extracted by the electrolysis of [2002] (a) bauxite (b) alumina (c) alumina mixed with molten cryolite (d) molten cryolite. The metal extracted by leaching with a cyanide is [2002] (a) Mg (b) Ag (c) Cu (d) Na. Which one of the following ores is best concentrated by froth-flotation method? [2004] (a) Galena (b) Cassiterite (c) Magnetite (d) Malachite During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’. These are [2005] (a) Fe and Ni (b) Ag and Au (c) Pb and Zn (d) Sn and Ag Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? [2008] (a) Metal sulphides are thermodynamically more stable than CS2 (b) CO2 is thermodynamically more stable than CS2 (c) Metal sulphides are less stable than the corresponding oxides (d) CO2 is more volatile than CS2

6.

Which method of purification is represented by the following equation ? [2012] 523K

Ti(s) + 2I 2 (g) ¾¾¾® 1700K

TiI 4 (g) ¾¾¾¾ ® Ti(s) + 2I 2 (g)

7.

8.

9.

(a) Zone refining

(b) Cupellation

(c) Polling

(d) Van Arkel

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is: [2014] (a) Ag

(b) Ca

(c) Cu

(d) Cr

In the context of the Hall - Heroult process for the extraction of Al, which of the following statements is false ? [JEE M 2015] 3+ (a) Al is reduced at the cathode to form Al (b) Na3AlF6 serves as the electrolyte (c) CO and CO2 are produced in this process (d) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity Which one of the following ores is best concentrated by froth floatation method? [JEEM 2016] (a) Galena (b) Malachite (c) Magnetite (d) Siderite

Answer Key 1 (c)

2 (b)

3 (c)

4 (b)

5 (c)

6 (d)

7 (b)

20

8 (b)

9 (a)

EBD_7764 C-96

1.

2.

Chemistry

(c) Pure aluminium can be obtained by electrolysis of a mixture containing alumina, crayolite and fluorspar in the ratio 20 : 24 : 20. The fusion temperature of this mixture is 900ºC and it is a good conductor of electricity. (b) Silver ore forms a soluble complex with NaCN from which silver is precipitated using scrap zinc.

6.

Zn

Ag 2 S + 2 NaCN ® Na[Ag (CN ) 2 ] ¾¾®

Na 2 [ Zn (CN ) 4 ] + Ag ¯ sodargento-cynanide (soluble)

3.

4. 5.

(c)

NOTE Galena is PbS and thus purified

by froth floatation method. Froath floatation method is used to concentrate sulphide ores. This method is based on th e preferential wetting properties with the froathing agent and water. (b) During the process of electrolytic refining Ag and Au are obtained as anode mud. (c)

7.

From this we find that on thermodynamic considerations CO2 is more stable than CS2 and the metal sulphides are more stable than corresponding oxides. In view of above the factor listed in choice (c) is incorrect and so is of no significance. Hence the correct answer is (c) (d) Van Arkel is a method in which heat treatment is used to purify metal in this process metals are converted into other metal compound for loosly coupled like as iodine to make metal iodide which are easily decomposed and give pure metal. The process is known as Van Arkel method. (b) On electrolysis of aqueous solution of s-block elements H2 gas discharge at cathode. At cathode: H 2O + e- ®

8.

(b)

9.

(a)

NOTE The reduction of metal sulphides

by car bon reduction process is not spontaneous because DG for such a process is positive. The reduction of metal oxide by carbon reduction process is spontaneous as DG for such a process is negative.

1 H 2 + OH 2

In the metallurgy of aluminium, purified Al2O3 is mixed with Na3AlF6 or CaF2 which lowers the melting point of the mix and brings conductivity. Froth floatation method is mainly applicable for sulphide ores. (1) Malachite ore : Cu(OH)2 . CuCO3 (2) Magnetite ore : Fe3O4 (3) Siderite ore : FeCO3 (4) Galena ore : PbS (Sulphide Ore)

The p-Block Elements (Group 15, 16, 17 & 18) 1.

2.

3.

4.

5.

6.

In XeF2, XeF4, XeF6 the number of lone pairs on Xe are respectively [2002] (a) 2, 3, 1 (b) 1, 2, 3 (c) 4, 1, 2 (d) 3, 2, 1. In case of nitrogen, NCl3 is possible but not NCl5 while in case of phosphorous, PCl 3 as well as PCl5 are possible. It is due to [2002] (a) availability of vacant d orbitals in P but not in N (b) lower electronegativity of P than N (c) lower tendency of H-bond formation in P than N (d) occurrence of P in solid while N in gaseous state at room temperature. Number of sigma bonds in P4O10 is [2002] (a) 6 (b) 7 (c) 17 (d) 16. Oxidation number of Cl in CaOCl2 (bleaching power) is: [2002] (a) zero, since it contains Cl2 (b) – 1, since it contains Cl – (c) + 1, since it contains ClO– (d) + 1 and – 1 since it contains ClO– and Cl– What may be expected to happen when phosphine gas is mixed with chlorine gas ? [2003] (a) PCl3 and HCl are formed and the mixture warms up (b) PCl5 and HCl are formed and the mixture cools down (c) PH3 . Cl2 is formed with warming up (d) The mixture only cools down Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that [2003] (a) oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas (b) strong affinity of HCl gas for moisture in air results in forming of droplets of liquid

21

solution which appears like a cloudy smoke. (c) due to strong affin ity for water, concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud. (d) concentrated hydrochloric acid emits strongly smelling HCl gas all the time. 7. Which one of the following substances has the highest proton affinity ? [2003] (a) H2S (b) NH3 (c) PH3 (d) H2O 8. Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ? [2004] (a) Hydration enthalpy (b) Ionization enthalpy (c) Electron affinity (d) Bond dissociation energy 9. Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction ? [2004] (a) Na2S2O3 is oxidised (b) CuI2 is formed (c) Cu2I2 is formed (d) Evolved I2 is reduced 10. Which one of the following statement regarding helium is incorrect ? [2004] (a) It is used to produce and sustain powerful superconducting magnets (b) It is used as a cryogenic agent for carrying out experiments at low temperatures (c) It is used to fill gas balloons instead of hydrogen because it is lighter and noninflammable (d) It is used in gas-cooled nuclear reactors 11. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is [2005]

EBD_7764 C-98

12.

13.

14.

15.

16.

17.

18.

(a) three (b) one (c) two (d) zero The correct order of the thermal stability of hydrogen halides (H–X) is [2005] (a) HI > HCl < HF > HBr (b) HCl< HF > HBr < HI (c) HF > HCl < HBr > HI (d) HI < HBr > HCl < HF Which of the following statements is true? [2006] (a) HClO4 is a weaker acid than HClO3 (b) HNO3 is a stronger acid than HNO2 (c) H3PO3 is a stronger acid than H2SO3 (d) In aqueous medium HF is a stronger acid than HCl The increasing order of the first ionization enthalpies of the elements B, P, S and F (Lowest first) is [2006] (a) B < P < S < F (b) B < S < P < F (c) F < S < P < B (d) P < S < B < F What products are expected from the disproportionation reaction of hypochlorous acid? [2006] (a) HCl and Cl2O (b) HCl and HClO3 (c) HClO3 and Cl2O (d) HClO2 and HClO4 Identify the incorrect statement among the following. [2007] (a) Br 2 reacts with hot and strong NaOH solution to give NaBr and H2O. (b) Ozone reacts with SO2 to give SO3. (c) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O. (d) Cl2 reacts with excess of NH3 to give N2 and HCl. Regular use of the following fertilizers increases the acidity of soil? [2007] (a) Ammonium sulphate (b) Potassium nitrate (c) Urea (d) Superphosphate of lime. Which one of the following reactions of xenon compounds is not feasible? [2009] (a) 3Xe F4 + 6H 2 O ¾¾ ® 2 Xe + Xe O3 +12HF +1.5O 2

(b)

2Xe F2 + 2H 2 O ¾¾ ® 2 Xe + 4HF + O 2

(c) (d)

Xe F6 + RbF ¾¾ ® Rb[Xe F7 ] Xe O3 + 6HF ¾¾ ® Xe F6 + 3H 2 O

Chemistry 19. Which of the following has maximum number of lone pairs associated with Xe ? [2011RS] (a) XeF4 (b) XeF6 (c) XeF2 (d) XeO3 20. The molecule having smallest bond angle is : [2012] (a) NCl3 (b) AsCl3 (c) SbCl3 (d) PCl3 21. Which among the following is the most reactive ? [JEE M 2015] (a) I2 (b) IC1 (c) Cl2 (d) Br2 22. Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason: The reaction between nitrogen and oxygen requires high temperature. [JEE M 2015] (a) The assertion is incorrect, but the reason is correct (b) Both the assertion and reason are incorrect (c) Both assertion and reason are correct, and the reason is the correct explanation for the assertion (d) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion 23. Which one has the highest boiling point ? [JEE M 2015] (a) Kr (b) Xe (c) He (d) Ne 24. The pair in which phosphorous atoms have a formal oxidation state of + 3 is : [JEE M 2016] (a) Orthophosphorous and hypophosphoric acids (b) Pyrophosphorous and pyrophosphoric acids (c) Orthophosphorous and pyrophosphorous acids (d) Pyrophosphorous and hypophosphoric acids 25. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces: [2016] (a) NO and N2O (b) NO2 and N2O (c) N2O and NO2 (d) NO2 and NO 26. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are : [2017] – – – (b) ClO2 and ClO3– (a) ClO and ClO3 – – (c) Cl and ClO (d) Cl– and ClO2–

C-99

The p-Block Elements (Group 15, 16, 17 & 18)

Answer Key 1 (d) 16 (d)

1.

2 (a) 17 (a)

3 (d) 18 (d)

4 (d) 19 (c)

5 (d) 20 (c)

6 (a) 21 (b)

7 (b) 22 (c)

8 (d) 23 (b)

9 (b) 24 (c)

10 (c) 25 (c)

11 (c) 26 (c)

12 (c)

13 (b)

14 (b)

15 (b)

are available. Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of d-orbitals

(d) In the formation of XeF 2 , sp 3 d hybridisation occurs which gives the molecule a trigonal bipyramidal structure.

5s

Xe

5p 5d Ground state configuration

O || P

••

••

F

•• F

3.

O

F

4.

P || O

F

x1

F

5.

F xe

2.

(a)

2 2 3 7N = 1s 2s 2p ;

F

6.

NOTE In phosphorous the 3d- orbitals

x2 = 2–1 = +1 now oxidation no. of another Cl is –1 as it is present as Cl–. (d) On mixing phosphine with chlorine gas PCl5 and HCl forms. The mixture cools down. PH3 + 4Cl2 –––––® PCl5 + 3HCl (a) 4HCl + O2 ® 2Cl2 + 2H2O air

F

2 2 6 2 3 15P = 1s 2s 2p 3s 3p

x2

hence oxidation no of Cl in OCl– is –2 + x2 = –1

In the formation of XeF 6 , sp 3 d 3 hybridization occurs which gives the molecule a pentagonal bipyramidal structure.

F

O

Ca (OCl) Cl F

F

O

(d) CaOCl2 –– or it can also be written as

xe

Excited state config.

P O

In the formation of XeF 4 , sp 3 d 2 hybridization occurs which gives the molecule an octahedral structure. F

O

P

||

||

O

Excited state configuration

O

O

(d)

7.

cloud of white fumes . .

(b) Among the given compounds, the N H 3 is most basic. Hence has highest proton affinity

EBD_7764 C-100 8. (d) The fluorine has low dissociation energy of F - F bond and reaction of atomic fluorine is exothermic in nature

9.

(b)

-1

0

4 KI +2CuSO4 ® I 2 + Cu 2 I 2 + 2 K 2SO 4 2+

0

+2.5

-1

I 2 + 2Na 2 S2 O3 ® Na 2 S4 O6 + 2 NaI

10.

In this CuI2 is not formed. (c) Helium is heavier than hydrogen although it is non-inflammable H

11.

|

(c) Hypophosphorous acid H - O - P ® O |

H

12.

Two H-atoms are attached to P atom. (c) The H–X bond strength decreases from HF to HI. i.e. HF > HCl > HBr > HI. Thus HF is most stable while HI is least stable. This is evident from their decomposition reaction: HF and HCl are stable upto 1473K, HBr decreases slightly and HI dissociates considerably at 713K. The decreasing stability of the hydrogen halide is also reflected in the values of dissociation energy of the H–X bond H-F

H - Cl

H - Br

Chemistry atom having fully or half filled stable orbitals. In this case P has a stable half filled electronic configuration hence its ionisation enthalapy is greater in comparision to S. Hence the correct order is B < S < PP>S>B NOTE On moving along a period

ionization enthalapy increases from left to right and decreases from top to bottom in a group. But this trend breaks up in case of

3l p

F

F Xe

F

XeF4 : F

2l p

C-101

The p-Block Elements (Group 15, 16, 17 & 18) 21.

F

F XeF6 :

Xe F

F

F

22.

F 1l p

23.

24.

Xe XeO3 : O

O

(b)

ICl Order of reactivity of halogens

Cl2 > Br2 > I2 But, the interhalogen compounds are generally more reactive than halogens (except F2), since the bond between two dissimilar electronegative elements is weaker than the bond between two similar atoms i.e, X – X (c) Nitrogen and oxgen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature. (b) Xe. As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases. (c) Phosphorous acid contain P in +3 oxidation state. Acid

O

1l p Hence XeF2 has maximum no. of lone pairs of electrons. 20. (c) All the members form volatile halides of the type AX3. All halides are pyramidal in shape. The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion. NCl3 PCl3 AsCl3 107° 94° 92°

Formula

Oxidation state of Phosphorus

Pyrophosphorous acid H4P2O 5 +3 Pyrophosphoric acid H4P2O 7 +5 Orthophosphorous acid H3PO 3 +3 Hypophosphoric acid H4P2O 6 +4 25. (c) Reaction of Zn with dil. HNO3 4Zn + 10HNO 3(dil) 4Zn(NO 3) 2 + 5H2O + N2O (Zn reacts differently with very dilute HNO3) Reaction of Zn with conc. HNO3 Zn + 4HNO 3 (conc.) Zn(NO 3) 2 + 2H2O + 2NO2

26. (c) Cl2 + NaOH ® NaCl + NaClO + H2O [cold and dilute]

EBD_7764 C-102

Chemistry

The d-and f-Block Elements 1.

2.

3.

Most common oxidation states of Ce (cerium) are [2002] (a) +2, +3 (b) +2, +4 (c) +3, +4 (d) +3, +5. Arrange Ce+3, La+3, Pm+3 and Yb+3 in increasing order of their ionic radii. [2002] +3 +3 +3 +3 (a) Yb < Pm < Ce < La (b) Ce+3 < Yb+3 < Pm+3 < La+3 (c) Yb+3 < Pm+3 < La+3 < Ce+3 (d) Pm+3 < La+3 < Ce+3 < Yb+3. Which of the following ions has the maximum magnetic moment? [2002] (a) Mn +2 (b) Fe+2 (c) Ti+2

4.

The most stable ion is (a) (c)

5.

6.

(d) Cr+2.

[Fe(OH)3]3[Fe(CN)6]3-

[2002] (b)

[Fe(Cl)6]3-

(d) [Fe(H2O)6]3+.

When KMnO4 acts as an oxidising agent and ultimately forms [MnO4]–2, MnO2, Mn2O3, Mn+2 then the number of electrons transferred in each case respectively is [2002] (a) 4, 3, 1, 5

(b) 1, 5, 3, 7

(c) 1, 3, 4, 5

(d) 3, 5, 7, 1.

The radius of La3+ (Atomic number of La = 57) is 1.06Å. Which one of the following given values will be closest to the radius of Lu3+ (Atomic number of Lu = 71) ? [2003]

7.

(a) 1.40 Å

(b) 1.06 Å

(c) 0.85 Å

(d) 1.60 Å

Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it ? [2003]

22

(a) In acidic solutions protons coordinate with ammonia molecules forming NH +4 ions and NH3 molecules are not available (b) In alkaline solutions insoluble Cu(OH)2 is precipitated which is soluble in excess of any alkali (c) Copper hydroxide is an amphoteric substance (d) In acidic solutions hydration protects copper ions 8. A red solid is insoluble in water. However it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is [2003] (a) HgI2 (b) HgO (c) Pb3O4 (d) (NH4)2Cr2O7 9. A reduction in atomic size with increase in atomic number is a characteristic of elements of [2003] (a) d-block (b) f-block (c) radioactive series (d) high atomic masses 10. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? [2003] (a)

Cr2 O 72- and H2O are formed

(b)

CrO 24- is reduced to +3 state of Cr

(c)

CrO 24- is oxidized to +7 state of Cr

(d) Cr3+ and Cr2O 72- are formed 11.

Which one of the following nitrates will leave behind a metal on strong heating ? [2003]

The d-and f-Block Elements

12.

13.

14.

15.

16.

(a) Copper nitrate (b) Manganese nitrate (c) Silver nitrate (d) Ferric nitrate Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them ? [2004] (a) (n – 1)d3 ns2 (b) (n – 1)d5 ns1 (c) (n – 1)d8 ns2 (d) (n – 1)d5 ns2 The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey power. This transformation is related to [2004] (a) a change in the partial pressure of oxygen in the air (b) a change in the crystalline structure of tin (c) an interaction with nitrogen of the air at very low temperature (d) an interaction with water vapour contained in the humid air Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by CN– ion towards metal species is [2004] (a) c, a (b) b, c (c) a, b (d) a, b, c Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect? [2004] (a) The +4 oxidation state of cerium is not known in solutions (b) The +3 oxidation state of cerium is more stable than the +4 oxidation state (c) The common oxidation states of cerium are +3 and +4 (d) Cerium (IV) acts as an oxidizing agent The correct order of magnetic moments (spin only values in B.M.) anong is [2004] (a)

[Fe(CN ) 6 ]4 - > [MnCl 4 ]2 - > [CoCl 4 ]2 -

(b) [MnCl 4 ]2- > [Fe(CN ) 6 ]4 - > [CoCl 4 ]2 (c)

[MnCl 4 ]2- > [CoCl 4 ]2 - > [Fe(CN ) 6 ]4 -

(d) [Fe(CN ) 6 ]4 - > [CoCl 4 ]2 - > [MnCl 4 ]2 (Atomic nos. : Mn = 25, Fe = 26, Co = 27)

C-103 17. Heating mixture of Cu2O and Cu2S will give [2005] (a) Cu2SO3 (b) CuO + CuS (c) Cu + SO3 (d) Cu + SO2 18. The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is: [2005] (a) + 3 (b) + 2 (c) + 6 (d) + 4

19. Calomel (Hg 2Cl 2 ) on reaction with ammonium hydroxide gives [2005] (a) HgO (b)

Hg 2O

(c)

NH 2 – Hg – Hg – Cl

(d) Hg NH 2 Cl 20. The lanthanide contraction is responsible for the fact that [2005] (a) Zr and Zn have the same oxidation state (b) Zr and Hf have about the same radius (c) Zr and Nb have similar oxidation state (d) Zr and Y have about the same radius 21. The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is [2005] (a)

d 5 (in strong ligand field)

(b)

d 3 (in weak as well as in strong fields)

(c)

d 4 (in weak ligand fields)

d 4 (in strong ligand fields) 22. Which of the following factors may be regarded as the main cause of lanthanide contraction? [2005] (a) Greater shielding of 5d electrons by 4f electrons (b) Poorer shielding of 5d electrons by 4f electrons (c) Effective shielding of one of 4f electrons by another in the subshell (d) Poor shielding of one of 4f electron by another in the subshell

(d)

EBD_7764 C-104 23. A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct? [2006] (a) MCl2 is more ionic than MCl4 (b) MCl2 is more easily hydrolysed than MCl4 (c) MCl2 is more volatile than MCl4 (d) MCl2 is more soluble in anhydrous ethanol than MCl4 24. Lanthanoid contraction is caused due to [2006] (a) the same effective nuclear charge from Ce to Lu (b) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge (c) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge (d) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge 25. The "spin-only" magnetic moment [in units of Bohr magneton, (µB )] of Ni 2+ in aqueous solution would be (At. No. Ni = 28) [2006] (a) 6 (b) 1.73 (c) 2.84 (d) 4.90 26. The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence [2007]

(a)

29.

30.

PbX 2 CH3– CH2X > (CH3)2 – CH.X > (CH3)3 – C–X (b) In Corey House synthesis of alkanes alkyl halide react with lithium dialkyl cuprate R 'X + LiR 2 Cu ¾¾ ® R '- R + RCu + LiX Br (c)

|

Alc. KOH

CH3 - CH - CH 2 - CH3 ¾¾¾¾¾ ®

CH3 - CH = CH - CH 3 + HBr The formation of 2-butene is in accordance to Saytzeff’s rule. The more substituted alkene is formed. (d) CH 3OH + C 6 H 5 MgBr ¾¾ ® CH 3O.MgBr + C6 H 6 (d) NH2 Na NO + HCl

N2+BF4

I

Nuclear substitution will not take place. 10. (c) In S N2 mechanism transition state is pentavelent. For bulky alkyl group it will have sterical hinderance and smaller alkyl group will favour the SN2 mechanism. So the decreasing order of reactivity of alkyl halides is RCH2X > R2CHX > R3CX 11. (d) SN2 reaction is favoured by small groups on the carbon atom attached to halogen. So, the order of reactivity is CH 3Cl > (CH 3 ) 2 CHCl > (CH 3 ) 3 CCl

> (C 2 H 5 ) 2 CHCl NOTE SN2 reaction is shown to maximum extent by

primary halides. The only primary halides given is CH3Cl so the correct answer is (d). 12. (a)

Me Me

+ BF3 + N2

Benzene diazonium tetrafluoroborate

(Balz-Schiemann reaction)

+Br

–

Me

Me + Br –

+

B

Be

D

ionisation

Br

HBF

¾¾®

Me

A Me

F

+

Br ionisation

Me

¾ ¾ ¾4 ®

0 - 5 º diazotisation

CH2CN

NaCN DM F

¾¾ ¾ ¾®

I

N2Cl ¾ ¾ ¾ ¾2¾ ¾¾®

C6H5

alc. KOH

ionisation

Me

Me +

+ Br

–

C

Since SN1 reactions involve the formation of carbocation as intermediate in the rate

EBD_7764 C-120

Chemistry

determining step, more is the stability of carbocation higher will be reactivity of alkyl halides towards SN1 route. Now we know that stability of carbocations follows the order : 3° > 2° > 1°, so SN1 reactivity should also follow the same order. 3° > 2° > 1° > Methyl (SN1 reactivity) Cl

13.

(c)

CH2

CH

CH2

CH3

Cl CH

CH

CH2Br CH3

A

Alcholic AgNO3

AgBr¯

COOH Acid (B)

15.

16.

2CH3CH 2 F + Hg 2Cl 2 19. (a) When tert -alkyl halides are used in Williamson synthesis elimination occurs rather than substitution resulting into formation of alkene. Here alkoxide ion abstract one of the b-hydrogen atom along with acting as a nucleophile. CH3

Oxidation COOH

2 - butyne

2CH3CH 2 Cl + Hg 2 F2 ¾¾ ®

Four monochloro derivatives are chiral. (d)

¾¾ ® CH3C º CCH3 + 6AgCl

CH3

CH3 (R + S)

14.

17. (c)

1, 1, 1-trichloroethane 18. (b) Alkyl fluorides are more conveniently prepared by heating suitable chloro – or bromo-alkanes with organic fluorides such as AsF3, SbF3, CoF2, AgF, Hg2F2 etc. This reaction is called Swarts reaction. CH 3Br + AgF ¾¾ ® CH 3 F + AgBr

CH3 ( R + S)

CH3

Cl | 2Cl - C - CH3 + 6Ag | Cl

CH3 D

CO

CH2

CH2

C

+–

CH3 + Na OCH3

CH3OH

Cl O

CO Phthalic Anhydride

2-Chloro-2-methylpentane

H CH3

CH OH (c) Acidic strength increases as the CH3CH2 C C CH3 + CH3OH + NaBr oxidation number of central atom 2-Methyl-pent-2-ene increases. Hence acidic strength order is (+7) (+5) (+3) (+1) 20. (b) Since SN1 reactions involve the formation HClO4 > HClO3 > HClO2 > HClO of carbocation as intermediate in the rate (b) Steric congestion around the carbon atom determining step, more is the stability of undergoing the inversion process will slow carbocation higher will be the reactivity of down the S N 2 reaction, h ence less alkyl halides towards SN1 route. congestion faster will the reaction. So, the Since stability of carbocation follows order. order is Å CH3Cl > (CH3)CH2 – Cl > (CH3)2CH – Cl > CH3 – CH2 – CH 2 (CH3)3CCl

C-121

Haloalkanes and Haloarenes Å

O

< CH3 – CH – CH2 – CH3 (d)

Å

tert - BuONa

Br

< p – H3CO – C6H4 – CH 2 II < I < III

Hence correct order is 21. (a) O

O tert - BuONa

(a)

¾¾ ¾ ¾ ¾ ® Br

O-tBu (fails to decolorise the colour of bromine) due to unsaturation

C6H5

(b)

C6H5

tert - BuONa

¾¾¾¾¾ ® Br (it decolorises bromine solution)

(c)

O

O tert - BuONa

Br

O ¾ ¾ ¾ ¾ ¾® (it decolorises bromine solution)

Products formed in option (2), (3) & (4) decolorises bromine solution due to presence of double bond. 22. (b) Elimination reaction is highly favoured if (a) Bulkier base is used (b) Higher temperature is used Hence in given reaction biomolecular elimination reaction provides major product.

Br

H C6H5

C6H5 H

¾ ¾ ¾ ¾ ¾®

t-BuOK 1, 2 elimination

H OtBu

(it decolorises bromine solution)

C6H5

C6H5 + t-BuOH + Br

EBD_7764 C-122

Chemistry

25

Alcohols, Phenols and Ethers 1.

2.

During dehydration of alcohols to alkenes by heating with conc. H2SO4 the initiation step is [2003] (a) formation of carbocation (b) elimination of water (c) formation of an ester (d) protonation of alcohol molecule Among the following compounds which can be dehydrated very easily is [2004]

CH 3

(a) CH 2 COOH OH

CH 3

OH

CH3

(a)

CH 2 COOH

(b)

|

CH3 CH 2 C CH 2 CH3

CH 3

|

OH

(c) CH(OH)COOH

OH |

(b)

CH3CH 2 CH 2CHCH3

(c)

CH 3CH 2 CH 2 CH 2 CH 2 OH

OH

CH 3 CH(OH)COOH

(d)

(d) CH3 CH 2 CHCH 2 CH 2 OH |

CH3

3.

4.

The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is [2005] (a) Pyridinium chloro-chromate (b) Chromic anhydride in glacial acetic acid (c) acidic dichromate (d) Acidic permanganate p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is [2005]

OH

5.

6.

HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give [ 2006] (a) BrCH2 – CH2 – OCH3 (b) H3C – CHBr – OCH3 (c) CH3CHO and CH3Br (d) BrCH2CHO and CH3OH Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is [ 2006]

(a)

CH3 | CH 3 - C HCH 2 OH

Alcohols, Phenols and Ethers

7.

(b) PhCHOHCH3 (c) CH3CH2CH(OH)CH2CH3 (d) C6H5CH2CH2OH The structure of the compound that gives a tribromo derivative on treatment with bromine water is [2006] CH 3

CH3 OH

(a)

(b)

OH

CH3

CH2OH

C-123 The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is [2009] (a) salicylaldehyde (b) salicylic acid (c) phthalic acid (d) benzoic acid 12. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is [2010] (a) 2-Butanol (b) 2- Methylpropan-2-ol (c) 2-Methylpropanol (d) 1- Butanol 13. The main product of the following reaction is

11.

conc.H SO

(c)

2 4¾ C6 H5CH 2CH(OH)CH(CH 3 )2 ¾¾¾¾¾ ®?

(d)

[2010]

OH 8.

–

+

O Na

OH + CHCl3 + NaOH

CHO

The electrophile involved in the above reaction is [2006]

(a)

H

Å

(b) formyl cation (CHO)

9.

H

(c) dichloromethyl cation (CHCl2) (d) dichlorocarbene (: CCl2) In the following sequence of reactions, P +I

Mg ether

CH3

C=C

CH3

C=C

CH(CH3)2 H

14. Consider thiol anion (RSQ ) and alkoxy anion

(ROQ ) . Which of the following statements is HCHO

2 ® A ¾¾¾ CH 3CH 2 OH ¾¾¾ ® B ¾¾¾¾ ®

correct ? (a)

H O

[2011RS]

RSQ is less basic but more nucleophilic

than ROQ

2 ®D C ¾¾¾

the compound D is

CH(CH3)2

H5C6CH2CH2 C = CH 2 H 3C

(d) C6H5

Å

H

C=C

(b) C6H5CH2 H (c)

(a) trichloromethyl anion (CCl3)

H 5C 6

[2007]

(a) propanal (b) butanal (c) n-butyl alcohol (d) n-propyl alcohol 10. Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives [2008] (a) 2, 4, 6-trinitrobenzene (b) o-nitrophenol (c) p-nitrophenol (d) nitrobenzene

(b)

RSQ is more basic and more nucleophilic than ROQ

(c)

RSQ is more basic but less nucleophilic

than ROQ (d)

RSQ is less basic and less nucleophilic

than ROQ 15. The correct order of acid strength of the following compounds : [2011RS]

EBD_7764 C-124

16.

Chemistry 19. The most suitable reagent for the conversion

(A) Phenol (B) p–Cresol (C) m–Nitrophenol (D) p–Nitrophenol (a) D > C > A > B (b) B > D > A > C (c) A > B > D > C (d) C > B > A > D Consider the following reaction :

of R - CH 2 - OH ® R - CHO is:

C2H5OH + H2SO4 ¾¾ ® Product Among the following, which one cannot be formed as a product under any conditions ? [2011RS] (a) Ethylene (b) Acetylene (c) Diethyl ether (d) Ethyl-hydr ogen sulphate 17. Arrange the following compounds in order of decreasing acidity : [2013]

OH

OH ;

OH ;

(a) KMnO4 (b) K2Cr2O7 (c) CrO3 (d) PCC (Pyridinium Chlorochromate) 20. Sodium phenoxide when heated with CO2 under pressure at 125ºC yields a product which on acetylation produces C [2014] ONa

H+ Ac2O

125° 5 Atm

+ CO 2 ¾¾¾¾ ® B ¾¾¾® C

The major product C would be

OCOCH3

OH ;

[2014]

COOH

(a)

; OH

Cl (I)

CH3 (II)

NO2 (III)

OCH3 (IV)

COCH3

(b)

(a) II > IV > I > III (b) I > II > III > IV (c) III > I > II > IV (d) IV > III > I > II 18. An unknown alochol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism : [2013] (a) secondary alcohol by SN1 (b) tertiary alcohol by SN1 (c) secondary alcohol by SN2 (d) tertiary alcohol by SN2

COCH3

OH COOCH3

(c)

OCOCH3 (d)

COOH

Answer Key 1 (d) 16 (b)

2 (a) 17 (c)

3 4 (a) (c) 18 19 (b) (d)

5 (b) 20 (a)

6 (b)

7 (c)

8 (d)

9 (d)

10 (b)

11 (b)

12 (b)

13 (a)

14 (a)

15 (a)

C-125

Alcohols, Phenols and Ethers

1.

(d) The dehydration of alcohol to form alkene occurs in following three step. Step (1) is initiation step. Step (1) Formation of protonated alcohol. + H CH3CH2 –– O H (Protonated ethanol)

+

CH3–CH2 – O –H + H

CH3 OH C OH

+

H Slow ––––––® H

+

2.

5.

H

OH

(b) Methyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction. HBr

® CH3 - CH - O - CH3 CH 2= CH - OCH3 ¾¾¾

Ethyl carbocation

Fast

|

Br 6.

(b) Only those alcohols which contain – CHOHCH3 group undergo haloform reaction. Among the given options only (b) contain this group, hence undergo haloform reaction.

7.

(c)

+

CH2 = CH 2+ H ethene (a) 3-methyl pentan-3-ol will be dehydrated most readily since it produces tertiary carbonium ion as intermediate.

NOTE OH group activates the benzene

nucleus and

CH3

OH

|

OH Br

CH3 - CH 2 - C - CH 2 - CH3 OH

CH3 |

¾¾¾ ® CH3 - CH 2 - C - CH 2 - CH3

8.

OH |

® CH3 - CH - CH = CH - CH3 ¾¾ O ||

and the electrophile is dichlorocarbene. Cl | H – C – Cl + NaOH | Cl

CH3 - C - CH = CH - CH3

Pyridiminum chloro-chromate (PCC) is specific for the conversion. CH3

4.

CH3 CHCl3 + NaOH

(c)

HCN

Reimer Tiemman reaction

OH

C=O OH

CH3

H

NOTE This is Riemer-Tiemann reaction

(d)

Å

(a)

¾¾ ¾ ¾ ¾® Br

CH3

H+

Br

Br2 / H 2O

|

3.

CH(OH) COOH

CN

+

Step (3)Elimination of a proton to form ethene H– CH 2––– CH2

HOH

Cyanohydrin

Step (2) Formation of carbocation CH3–CH2–– O

CH3

••

Cl – C – Cl + NaCl + H2O dichlorocarbene

a-elimination

9.

(d)

P +I

2 ® CH CH I CH3CH 2 OH ¾¾¾ 3 2

A

Mg Ether

¾¾¾®

CH3CH 2 MgI (B)

EBD_7764 C-126

Chemistry

CH 2 CH3 | H - C - OMgI | H (C)

HCHO

¾¾¾¾ ®

Mechanism CH3 | Step 1 : CH3 — C — OH + H - Cl |

CH3

2 Methyl Propan-2-ol

H2 O

¾¾¾ ®

CH 2 CH3 | H - C - OH | H (D)

Step 2 : Step 3 :

(b) Phenol on reaction with conc. H2SO4 gives a mixture of o- and p- products (i.e., –SO3H group, occupies o-, p- position). At room temperature o-product is more stable, which on treatment with conc. HNO3 will yield o-nitrophenol.

OH Conc.H2SO4

OH Conc.H SO

OH

SO3H +

o-

SO3H pAt room temperature o- product is more stable OH SO3H

OH

∗

(CH3 )3 C ∗ Cl ‡ˆˆ (CH3 )3 C , Cl t,Butylchloride

Conc.H SO

2 4¾ C6H5 ,CH2 ,CH, CH ,CH3 ¾¾¾¾¾ ↑

|

OH

11.

(b)

12.

(b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl2 (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation.

COOH

|

OH CH3

Conc. HNO 3

NaOH CO2

, ˆˆ†

13. (a) Whenever dehydration can produce two different alkenes, major product is formed according to Saytzeff rule i.e. more substituted alkene (alkene having lesser number of hydrogen atoms on the two doubly bonded carbon atoms) is the major product. Such reactions which can produce two or more structural isomers but one of them in greater amounts than the other are called regioselective ; in case a reaction is 100% regioselective, it is termed as regiospecific. In addition to being regioselective, alcohol dehydrations are stereoselective (a reaction in which a single starting material can yield two or more stereoisomeric products, but gives one of them in greater amount than any other).

NO2

o- nitrophenol Hence (b) is the correct answer. OH

∗

ˆˆ† (CH 3 )3 C∗ ∗ H 2 O (CH3 )3 C , OH 2 ‡ˆˆ 3° Carbocation

n - propyl alcohol

10.

∗

ˆˆ† (CH3 )3 C — OH 2 ∗ Cl, ‡ˆˆ

H

H C=C

C6H5

CH(CH3)2 cis (minor)

H +

CH(CH3)2 C=C

C6H5

H trans (major)

14. (a) On moving down a group, the basicity & nucleophilicity are inversely related, i.e. nucleophilicity increases while basicity decreases. i.e RSQ is more nucleophilic but less basic than ROQ . This opposite

C-127 – CH3, – OCH3 decreases acidity. hence the correct order of acidity will be

Alcohols, Phenols and Ethers

behaviour is because of the fact that basicity and nucleophilicity depends upon different factors. Basicity is directly related to the strength of the H–element bond, while nucleophilicity is indirectly related to the electronegativity of the atom to which proton is attached. OH

15.

OH

OH

(a)

(A)

NO2

CH3 +I

–I

effect (B)

effect (C)

OH

OH

OH >

OH >

OH >

NO2 Cl CH3 OCH3 III I II IV (–M, –I) (–I > +M) (+I, + HC ) (+ M) 18. (b) Reaction of alcohols with Lucas reagent proceeds through carbocation formation, SN1 mechanism. Further 3° carbocations (from tertiary alcohols) are highly stable thus reaction proceeds through SN1 mechanism. 19. (d) An excellent reagent for oxidation of 1° alcohols to aldehydes is PCC. PCC

R - CH 2 - OH ¾¾¾ ® R - CHO

ONa

NO2 – M, – I effect (D) Electron withdrawing substituents increases the acidity of phenols; while electron releasing substituents decreases acidity. Since the + I effect is maximum in ortho position, followed by meta and least in para, thus the correct order of acidity will D > C >A> B 16. (b) C2 H5 - OH + H2 SO4

® + CO2 ¾¾

20. (a)

433K

413 K

OH H SO

2 4® ¾¾¾¾

COONa

OH COOH Salicylic acid (CH 3 CO) 2 O

CH 2 = CH 2 ethylene

CH3 - CH 2 - O - CH2 - CH3 diethyl ether

383 K

Sodium Phenoxide

CH3CH 2 HSO4 + H 2 O

O || O - C - CH3

ethyl hydrogen sulphate

17.

Acetylene is not formed under any conditions. (c) Electron withdrawing substituents like –NO2, –Cl increase the acidity of phenol while electron releasing substituents like

COOH Aspirin (Acetyl Salicylate)

+ CH3COOH

EBD_7764 C-128

Chemistry

26

Aldehydes, Ketones and Carboxylic Acids 1.

Cl 2 KOH ¾¾® A ¾alc. ¾ ¾¾® B. CH3CH2COOH ¾red P

2.

What is B? [2002] (a) CH3CH2COCl (b) CH3CH2CHO (c) CH2=CHCOOH (d) ClCH2CH2COOH. On vigorous oxidation by permanganate solution. (CH3)2C = CH - CH2 - CHO gives [2002]

(a)

CH OH | | CH3 - C - CH - CH 2CH 3 | CH

(b)

CH3 CH3

(c)

CH3

4.

5.

6. COOH + CH3CH2COOH

CH – OH + CH2CH2CH2OH

CH3

(d) .

CH3 CH3

3.

Picric acid is:

[2002]

COOH

COOH

(a)

8.

(b) NO2

OH COOH

OH (c)

7.

C = O + CH2CH2CHO

O2N

NO2

9.

pK a value?

(d) NH2

NO2

When CH2 = CH — COOH is reduced with LiAlH4, the compound obtained will be [2003] (a) CH2 = CH — CH2OH (b) CH3 — CH2 — CH2OH (c) CH3 — CH2 — CHO (d) CH3 — CH2 — COOH On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is [2004] (a) CH3COCl+ C2H5OH + NaOH (b) CH3COONa + C2H5OH (c) CH3COOC2H5 + NaCl (d) CH3Cl + C2H5COONa Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl gives [2004] (a) 2-methyl-2propanol (b) acetamide (c) acetone (d) acetyl iodide Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon? [2004] (a) Acetamide (b) Acetic acid (c) Ethyl acetate (d) Butan-2-one Which one of the following undergoes reaction with 50% sodium hyroxide solution to give the corresponding alcohol and acid? [2004] (a) Butanal (b) Benzaldehyde (c) Phenol (d) Benzoic acid Among the following acids which has the lowest (a)

CH 3CH 2COOH

(b)

(CH 3 ) 2 CH - COOH

[2005]

Aldehydes, Ketones and Carboxylic Acids

(c) HCOOH (d)

CH 3COOH

10.

Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known [2005] (a) an amine (b) an imine (c) an anemine (d) a Schiff’s base 11. The increasing order of the rate of HCN addition to compound A – D is [2006] (A) HCHO (B) CH3COCH3 (C) PhCOCH3 (D) PhCOPh (a) D < C < B < A (b) C < D < B < A (c) A < B < C < D (d) D < B < C < A 12. The correct order of increasing acid strenght of the compounds [2006] (A) CH3CO2H (B) MeOCH2CO2H (C) CF3CO2H

(D)

Me Me

CO2H

is (a) D < A < B < C (b) A < D < B < C (c) B < D < A < C (d) D < A < C < B 13. A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was : [2009] (a) HCHO (b) CH3COCH3 (c) CH3COOH (d) CH3OH 14. Which of the following on heating with aqueous KOH, produces acetaldehyde? [2009] (a) CH3CH2Cl (b) CH2ClCH2Cl (c) CH3CHCl2 (d) CH3COCl 15. In Cannizzaro reaction given below Π .. : ΝΓ 2PhCHO ¾¾¾ ↑ PhCH 2 OH + PhCOΠ 2 the slowest step is : [2009] (a) the transfer of hydride to the carbonyl group (b) the abstraction of proton from the carboxylic group (c) the deprotonation of Ph CH2OH

C-129 16. Iodoform can be prepared from all except: [2012] (a) Ethyl methyl ketone (b) Isopropyl alcohol (c) 3-Methyl 2-butanone (d) Isobutyl alcohol 17. In the given transformation, which of the following is the most appropriate reagent ? [2012]

CH

COCH3

HO Reagent

¾¾¾¾® CH

CH

CH2

CH3

HO (a)

Q

NH2 NH2 , OH

(b) Zn – Hg/ HCl

(c) Na, Liq NH3 (d) NaBH4 18. An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to given CH3CH2NH2. A is : [2013] (a) CH3COOH (b) CH3CH2CH2COOH (c)

CH 3 - CH - COOH | CH3

(d) CH3CH2COOH 19. In the reaction, LiAlH

4® A CH 3COOH ¾¾¾¾

PCl

Alc.KOH

5 ® B ¾¾¾¾® C, ¾¾¾

the product C is: (a) Acetaldehyde (c) Ethylene

[2014] (b) Acetylene (d) Acetyl chloride

20. The correct sequence of reagents for the following conversion will be : O HO CH3

Π

(d) the attack of : OH at the carboxyl group

CH

CHO

HO–CH3 CH3

EBD_7764 C-130

Chemistry

(a) [Ag(NH3)2]+ OH–, H+/CH3OH, CH3MgBr (b) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+ OH– (c) CH3MgBr, [Ag(NH3)2]+ OH–, H+/CH3OH (d) [Ag(NH3)2]+ OH–, CH3MgBr, H+/CH3OH 21. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is : (a) C6H5COONa (b) HCOONa (c) CH3COONa (d) Na2C2O4 22. The major product obtained in the following reaction is :

OH CHO

(a)

COOH

OH (b)

CHO CHO CHO

(c)

O

COOH

O

CHO

(d)

DIBAL–H

CHO COOH

Answer Key 1 (c) 16 (d)

1.

2 (b) 17 (a)

3 (c) 18 (d)

4 (a) 19 (c)

5 (c) 20 (a)

6 (a) 21 (d)

7 (d) 22 (b)

8 (b)

Cl2 ¾® CH3CHClCOOH (c) CH3CH2COOH ¾¾ red P

9 (c)

5.

¾¾ ¾¾® CH 2 = CHCOOH

2.

3. 4.

11 (a)

12 (a)

13 (c)

14 (c)

(c) There is no reaction hence the resultant mixture contains CH3 COOC2H5 + NaCl.

Acrylic acid

(b) Aldehydic group gets oxidised to carboxylic group. Double bond breaks and carbon gets oxidised to carboxylic group. (c) 2,4,6-Trinitrophenol is also known as picric acid. (a) LiAlH4 can reduce COOH group and not the double bond. LiAlH

4 ® CH CH 2 = CH - COOH ¾¾¾¾ 2

15 (a)

CH2 = CH - CH 2OH

alc.KOH

- HCl

10 (c)

O

6.

(a)

||

(i)CH MgI (ii)Saturated NH 4Cl

3 CH 3 - C- Br ¾¾¾¾¾¾¾¾ ®

CH3 |

CH3 – C – OH

| CH3 2-methyl-2-propanol

C-131 (Me)2CHCOOH < CH3COOH < MeOCH2COOH < CF3COOH

Aldehydes, Ketones and Carboxylic Acids 7. (d) It is Clemmensen’s reduction

O ||

Zn - Hg CH3 - C - CH 2 - CH3 ¾¾¾¾®

[

Conc.HCl

Butane - 2 -one

Zn Hg

¾¾¾¾® CH3CH 2 - CH 2 CH3 Conc.HCl (Butane)

8.

(b) This reaction is known as cannizzaro’s reaction. In this reaction benzaldehyde in presence of 50%. NaOH undergoes disproportionation reaction and form one mol of Benzyl alcohol (Red. product) and one mole of sod. benzoate (ox. product)

NOTE Electron withdrawing groups

increase the acid strength and electron donating groups decrease the acid strength.] 13. (c) Fruity smell is due to ester formation which is formed between ethanol and acid. Conc. H 2SO 4 CH3COOH + C2H5OH ¾¾¾¾¾¾ ® CH3COOC2H5 + H2O aq.KOH 14. (c) CH 3CHCl 2 ¾¾¾¾ ® CH 3CH(OH) 2 -H O

CH2OH

CHO

2 ® CH CHO ¾¾¾¾ 3

50% NaOH

O +

COO Na

–

+

9.

10.

15. (a)

||

fast

ˆˆˆ† Ph – C –H + OH – ‡ˆˆˆ

– O Ph – C – H OH

(c) pKa = –log Ka; HCOOH is the strongest acid and hence it has the highest Ka or lowest pKa value. (c)

||

enamine

11.

(a)

NOTE Addition of HCN to carbonyl

compounds is nucleophilic addition reaction. The order of reactivity of carbonyl compounds is Aldehydes (smaller to higher) Ketones (smaller to higher), Then HCHO > CH3COCH3 > PhCOCH3 > PhCOPh NOTE The lower reactivity of Ketones

12.

|

is due to presence of two alkyl group which shows +I effect. The reactivity of Ketones decreases as the size of alkyl group increases. (a) The correct order of increasing acid strength

|

OH

O N (CH3)2

|

Ph – C + Ph – C –H

N (CH3)2 –H2O

O–

O

OH O + HN(CH3)2

O Ph – C – H slow

+

H exchange ¾¾¾¾¾¾ ® Ph fast

||

H

OH |

– C + Ph – C –H |

O–

|

H

16. (d) Iodoform test is given by methyl ketones, acetaldehyde and methyl secondary alcohols. CH3

CH

CH2

OH

CH3

isobutyl alcohol is a primary alcohol hence does'nt give positive iodoform test. 17. (a) Aldehydes and ketones can be reduced to hydrocarbons by the action (i) of amalgamated zinc and concentrated hydrochloric acid (Clemmensen reduction), or (b) of hydrazine (NH2NH2) and a strong base like NaOH, KOH or potassium tertbutoxide in a high-boiling alcohol like

EBD_7764 C-132

Chemistry

ethylene glycol or triethylene glycol (WolfKishner reduction) CH

CH

COCH3

O [Ag(NH3)2]OH

20. (a)

HO NH 2NH 2/OH

Tollens reagent

–

CHO

Wolf-kishner Reduction

CH

CH

CH2

O

CO2H H+/CH3OH (esterification)

CH3

O

HO

–OH group and alkene are acid-sensitive groups so clemmensen reduction can not be used.Acid sensitive substrate should be reacted in the Wolf-Kishner reduction which utilise strongly basic conditions. 18.

(d)

O

NH

D

(I)

II

Br KOH,(III)

3 ® B ¾¾ 2 ® CH CH NH ® C ¾¾¾¾¾ A ¾¾¾ 3 2 2

CH3

H3C – C – CH3 OH

O CH3CH 2 , C, OH ¾¾¾ ↑ CH3CH 2COO, NH∗ 4 NH3

(A)

(B) Χ

¾¾↑ CH 3CH 2 CONH 2

(C) KOH

Br2

CH 3CH 2 NH 2

19.

(c)

OCH3 CH3MgBr

Reaction (III) is a Hofmann bromamide reaction. Now formation of CH3CH2NH2 is possible only from a compound CH3CH2CONH2(C) which can be obtained from the compound CH3CH2COO– NH+4 (B). Thus (A) should be CH3CH2COOH P

C

LiA1H

4 ® CH CH OH CH3 COOH ¾¾¾¾ 2 3

(A)

PCl5

CH3CH2Cl (B)

21. (d) Na2C2O4 + H2SO4 ® Na2SO4 + CO­ + CO2­ + H2O 'x'

(conc.)

Na2C2O4 + CaCl2 ® CaC2O4¯ + 2NaCl 'x'

(white ppt.)

5CaC2O4 ¯ + 2KMnO4 + 8H2SO4 (purple)

K2SO4 + 5CaSO4 + 2MnSO4 + 10CO2 + 8H2O (colourless)

22. (b) DIBAL-H is an electrophilic reducing agent. It reduces both ester and carboxylic group into an aldehyde at low temperature. O

O

OH

O

Alc. KOH

CH2 = CH2 (C)

Hence the product (C) is ethylene.

H

DIBAL - H

¾ ¾ ¾ ¾® CO2H

CHO

27

Amines 1.

2.

When primary amine reacts with chloroform in ethanolic KOH then the product is [2002] (a) an isocyanide (b) an aldehyde (c) a cyanide (d) an alcohol. The reaction of chloroform with alcoholic KOH and p-toluidine forms [ 2003]

6.

7.

3.

4.

5.

(a) H3C

N2Cl

(b) H3C

NHCHCl2

(c) H3C

NC

(d) H3C

CN

The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2NH is [ 2003] (a) (CH3)2NH < NH3 < CH3NH2 (b) NH3 < CH3NH2 < (CH3)2NH (c) CH3NH2 < (CH3)2NH < NH3 (d) CH3NH2 < NH3 < (CH3)2NH Ethyl isocyanide on hydrolysis in acidic medium generates [2003] (a) propanoic acid and ammonium salt (b) ethanoic acid and ammonium salt (c) methylamine salt and ethanoic acid (d) ethylamine salt and methanoic acid Which one of the following methods is neither meant for the synthesis nor for separation of amines? [2005]

(a) Curtius reaction (b) Wurtz reaction (c) Hofmann method (d) Hinsberg method Amongst the following the most basic compound is [2005] (a) p -nitroaniline (b) acetanilide (c) aniline (d) benzylamine An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is [2005] (a)

CH 3CH 2CONH 2 (b) ( NH 2 ) 2 CO

(c)

CH 3CONH 2

(d) CH 3 NCO

8.

Which one of the following is the strongest base in aqueous solution ? [2007] (a) Methylamine (b) Trimethylamine (c) Aniline (d) Dimethylamine 9. In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH ® (A) + (B) + 3H2O, the compounds (A) and (B) are respectively [2007] (a) C2H5NC and 3KCl (b) C2H5CN and 3KCl (c) CH3CH2CONH2 and 3KCl (d) C2H5NC and K2CO3. 10. In the chemical reactions, NH 2 NaNO2 HCl, 278 K

A

HBF 4

B

EBD_7764 C-134

11.

Chemistry

the compounds ‘A’ and ‘B’ respectively are [2010] (a) nitrobenzene and fluorobenzene (b) phenol and benzene (c) benzene diazonium chloride and fluorobenzene (d) nitrobenzene and chlorobenzene In the chemical reactions : [2011RS] NH 2 NaNO

NaNO /HCl

the product E is : CN

(a) CH3

CH3

CuCN

2 ® A ¾¾¾¾ ¾¾¾¾¾ ® B, HCl, 278K D

the compounds A and B respectively are : (a) Benzene diazonium chloride and benzonitrile (b) Nitrobenzene and chlorobenzene (c) Phenol and bromobenzene (d) Fluorobenzene and phenol 12. A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is : [2013] (a) 2 (b) 5 (c) 4 (d) 6 13. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is: [2014] (a) an alkanol (b) an alkanediol (c) an alkyl cyanide (d) an alkyl isocyanide 14. Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value? [2014] (a) (CH3)2NH (b) CH3NH2 (c) (CH3)3N (d) C6H5NH2 15.

In the reaction NH2

CuCN/KCN

2 ¾¾¾¾¾ ® D ¾¾¾¾¾ ® E + N2 0–5° C D

(b) COOH

(c) CH3

CH3

(d) H3C

16. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br 2 used per mole of amine produced are : [JEE M 2016] (a) Two moles of NaOH and two moles of Br 2. (b) Four moles of NaOH and one mole of Br 2. (c) One mole of NaOH and one mole of Br 2. (d) Four moles of NaOH and two moles of Br 2. 17. Which of the following compounds will form significant amount of meta product during mononitration reaction ? [2017] OH OCOCH3

[JEE M 2015] (a)

(b) NH2

(c) CH3

NHCOCH3

(d)

C-135

Amines

Answer Key 1 (a) 16 (b)

1.

2 (c) 17 (c)

3 (b)

4 (d)

5 (b)

6 (d)

7 (b)

8 (d)

9 (a)

10 (c)

11 (a)

12 (b)

13 (d)

(a) C2H5NH2 + CHCl3 + 3KOH ® C2H5N º C + 3KCl + 3HCl

6.

(d) Benzylamine

14 (a)

15 (a)

.. CH2NH2

is most

(Ethyl isocyanide)

basic. In others the basic character is suppressed due to Resonan ce (see applications of resonance).

NH2 2.

(c)

+CHCl3+ 3KOH 7.

CH3

(a)

N=C +3KCl+3H2 O CH3

3.

(b) The alkyl groups are electron releasing group (+ I), thus increases the electron density around the nitrogen thereby increasing the availability of the lone pair of electrons to proton or lewis acid and making the amine more basic. Hence more the no. of alkyl group more basic is the amine. Therefore the correct order is NH3 < CH3NH2 < (CH3)2 NH

4.

(d) Ethyl isocyanide on hydrolysis form primary amines. +

H = C + H2 O ¾¾¾ CH3CH2 N ® ®

CH3CH2NH2 + HCOOH Therefore it gives only one mono chloroalkane. 5.

(b)

(b) Wurtz reaction is for the preparation of hydrocarbons from alkyl halide RX + 2 Na + XR ¾ ¾® R – R + 2 NaX

(b)

(c)

(d)

C

20%

20/12 = 1.66

1.66 / 1.66 = 1

H

6.67%

6.67 / 1 = 6.67

6.67 / 1.66 = 4.16

N

46.67%

46.67/14 = 3.33

3.33 / 1.66 = 2.02

O

26.64%

26.64 / 16 = 1.66 1.66 / 1.66 = 1.0

The compound is CH4N2O Empirical weight = 60; Mol. wt. = 60; \n =

60 =1 60

O ||

Molecular formula = CH4N2O; NH 2 - C - NH 2 On heating urea loses ammonia to give Biuret 2NH 2CONH 2 ¾¾ ® H 2 NCO.NH.CONH 2 + NH 3

Biuret with alkaline CuSO4 gives violet colour. Test for –CONH– group. 8. (d)

NOTE Aromatic amines are less basic than

aliphatic amines. Among aliphatic amines the order of basicity is 2° > 1° > 3°. The electron density is decreased in 3° amine due to crowding of alkyl group over N atom which makes the approach and bonding by a proton

EBD_7764 C-136

Chemistry

relatively difficult. Therefore the basicity decreases. Further Phenyl group show – I effect, thus decreases the electron density on nitrogen atom and hence the basicity. \ dimethylamine (2° aliphatic amine) is strongest base among given choices. \ The correct order of basic strength is Dimethylamine > Methyl amine > Trimethyl amine > Aniline. 9. (a) This is carbylamine reaction. CH3CH2NH2 + CHCl3 + 3KOH

¾¾ ® C2H5NC + 3KCl + 3H2O 10. (c) Primary aromatic amines react with nitrous acid to yield arene diazonium salts. cold + NaNO2 + 2HX ¾¾¾ ® 1° Aromatic amine Ar—N = N+X– + NaX + 2H2O Arene diazonium salt The diazonium group can be replaced by fluorine by treating the diazonium salt with fluoroboric acid (HBF4). The precipitated diazonium fluoroborate is isolated, dried and heated until decomposition occurs to yield the aryl fluoride. This reaction is known as Balz-Schiemann reaction.

12. (b)

O P R - NH - C - CH3 144244 3 Mol.mass =58

Now since the molecular mass increases by 42 unit as a result of the reaction of one mole of CH3COCl with one-NH2 group and the given increase in mass is 210. Hence the number of –NH2 groups is = 210/42 = 5. 13. (d) R – CH2 – NH2 + CHCl3 + 3KOH (alc) Carbyl amine reaction

¾¾ ® R – CH2 – NC + 3KCl + 3H2O Alkyl isocynide 14. (a) Arylamines are less basic than alkyl amines and even ammonia. This is due to resonance. In aryl amines the lone pair of electrons on N is partly shared with the ring and is thus less available for sharing with a proton. In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom and thus also increases the ability of amine for protonation. Hence more the no. of alkyl groups higher should be the basicity of amine. But a slight discrepancy occurs in case of trimethyl amines due to steric effect. Hence the correct order is

HBF4 heat Ar—N2+X– ¾¾¾® Ar—N2+BF4–¯ ¾¾¾ ®

Ar—F + BF3 + N2

(a)

N +2 Cl-

NH 2 NaNO2

CuCN

(CH 3 )2 NH > CH3 NH 2

HCl, 278K Diazotization

- HCl

Mol.Mass=16

ArNH2

11.

R - NH 2 {

O P + CH3 - C - Cl ¾¾¾®

> (CH3 )3 N > C6 H5 NH 2

Benzene diazonium chloride

15. (a)

(A) C ºN

+

NH2 0 - 5° C

CH3

CN

¾ CuCN/KCN ¾ ¾ ¾ ¾®

¾ NaNO ¾ ¾ 2¾/HCl ¾¾ ® Benzonitrile (B) Sandmayer reaction

–

N = NCl

D

CH3

CH3

Amines 16. (b) 4 moles of NaOH and one mole of Br2 is required during production of one mole of amine during Hoffmann's bromamide degradation reaction.

O || R–C–NH2 + Br2 + 4NaOH ® R–NH2 + K2CO3 + 2NaBr + 2H2O 17.

(c) Nitration takes place in presence of concentrated HNO3 + concentrated H2SO4

C-137 In strongly acidic nitration medium, the amine is converted into anilinium ion (– NH3+); substitution is thus controlled not by – NH2 group but by – NH3+ group which, because of its positive charge, directs the entering group to the metaposition instead of ortho, and para. Å

NH3

: NH2 Conc. HNO

3 ¾ Conc. ¾ ¾H¾SO¾® 2

– NH2 gp : o,p -directing

4

– NH+3 gp : m-directing

EBD_7764 C-138

Chemistry

28

Biomolecules 1.

2.

3.

4.

5.

6.

7.

RNA is different from DNA because RNA contains [2002] (a) ribose sugar and thymine (b) ribose sugar and uracil (c) deoxyribose sugar and thymine (d) deoxyribose sugar and uracil. Complete hydrolysis of cellulose gives [2003] (a) D-ribose (b) D-glucose (c) L-glucose (d) D-fructose The reason for double helical structure of DNA is operation of [2003] (a) dipole-dipole interaction (b) hydrogen bonding (c) electrostatic attractions (d) van der Waals’ forces Which base is present in RNA but not in DNA ? (a) Guanine (b) Cytosine [2004] (c) Uracil (d) Thymine Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories ? [2004] (a) An enzyme (b) A hormone (c) A co-enzyme (d) An antibiotic Which of the following is a polyamide? [2005] (a) Bakelite (b) Terylene (c) Nylon-66 (d) Teflon In both DNA and RNA, heterocylic base and phosphate ester linkages are at – [2005] (a)

C5' and C1' respectively of the sugar

molecule

(b)

C1' and C5' respectively of the sugar

molecule (c)

C '2 and C5' respectively of the sugar

molecule (d)

C5' and C '2 respectively of the sugar

molecule The term anomers of glucose refers to [2006] (a) enantiomers of glucose (b) isomers of glucose that differ in configuration at carbon one (C-1) (c) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4) (d) a mixture of (D)-glucose and (L)-glucose 9. The pyrimidine bases present in DNA are [2006] (a) cytosine and thymine (b) cytosine and uracil (c) cytosine and adenine (d) cytosine and guanine 10. The secondary structure of a protein refers to [2007] (a) fixed configuration of the polypeptide backbone (b) a– helical backbone (c) hydrophobic interactions (d) sequence of a– amino acids. 11. a - D-(+)-glucose and b-D-(+)-glucose are [2008] (a) conformers (b) epimers (c) anomers (d) enatiomers 8.

C-139

Biomolecules

12.

13.

14.

15.

16.

17.

The two functional groups present in a typical carbohydrate are: [2009] (a) – CHO and – COOH (b) > C = O and – OH (c) – OH and – CHO (d) – OH and – COOH Biuret test is not given by [2010] (a) carbohydrates (b) polypeptides (c) urea (d) proteins Which of the following compounds can be detected by Molisch's Test ? [2012] (a) Nitro compounds (b) Sugars (c) Amines (d) Primary alcohols Which one of the following statements is correct? [2012] (a) All amino acids except lysine are optically active (b) All amino acids are optically active (c) All amino acids except glycine are optically active (d) All amino acids except glutamic acids are optically active Synthesis of each molecule of glucose in photosynthesis involves : [2013] (a) 18 molecules of ATP (b) 10 molecules of ATP (c) 8 molecules of ATP (d) 6 molecules of ATP Which one of the following bases is not present in DNA? [2014]

(a) Quinoline (b) Adenine (c) Cytosine (d) Thymine 18. Which of the vitamins given below is water soluble ? [2015] (a) Vitamin E (b) Vitamin K (c) Vitamin C (d) Vitamin D 19. Thiol group is present in : [2016] (a) Cysteine (b) Methionine (c) Cytosine (d) Cystine 20. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

HOH2C

CH2OH

O

(a)

HO OCOCH

3

OH HOH2C

CH2OH

O

(b)

HO OH HOH2C

CH2OH

O

(c)

HO OCH

3

OH HOH2C O CH2OCH3

(d)

OH OH

OH

Answer Key 1 (b) 16 (b)

2 (b) 16 (a)

3 (b) 17 (a)

4 (c) 18 (c)

5 (b) 19 (a)

6 (c) 20 (a)

7 (b)

8 (b)

9 (a)

10 (b)

11 (c)

12 (c)

13 (a)

14 (b)

15 (c)

EBD_7764 C-140

1.

2.

Chemistry

(b) In RNA, the sugar is D–ribose and base is uracil where as in DNA, the sugar is D-2 deoxyribose and the nitrogenous base is thymine. (b) Cellulose is a linear polymer of b –D– glucose in which C1 of one glucose unit is connected to C4 of the other through b –D glucosidic linkage. It does not undergo hydrolysis easily. However on heating with dilute H2SO4 under pressure. It does undergo hydrolysis to give only D– glucose. H+

(C6 H10O5 )n+nH 2O ¾¾® nC6 H12O6 D-Glucose

3.

(b) DNA consists of two polynucleotide chains, each chain forms a right handed spiral with ten bases in one turn of the spiral. The two chains coil to double helix and run in opposite direction held together by hydrogen bonding.

4.

(c) RNA contains cytosine and uracil as pyrimidine bases while DNA has cytosine and thymine. Both have the same purine bases i.e., Guanine and adenine. (b) Insulin is a biochemically active peptide harmone secreted by pancreas. (c) Nylon is a general name for all synthetic fibres forming polyamides. (b) In DNA and RNA heterocyclic base and phosphate ester are at C 1' and C5' respectively of the sugar molecule.

5. 6. 7.

HO | 5 HO – P – O – C H 2 || O H

8.

N 4¢ C

O H H | | C3––– C2 | | OH OH

N

N N

C¢ H

(b) Cyclization of the open chain structure of D-(+)-glucose has created a new stereocenter at C 1 which explains the existence of two cyclic forms of D-(+)-glucose, namely a– and b–. These two cyclic forms are diasteromers, such diastereomers which differ only in the configuration of chiral carbon developed on hemiacetal formation (it is C1 in glucose and C2 in fructose) are called anomers and the hemiacetal carbon (C1 or C2) is called the anomeric carbon.

C-141

Biomolecules

9. 10.

11.

(a) The pyrimidine bases present in DNA are cytosine and thymine. (b) The secondary structure of a protein refers to the shape in which a long peptide chain can exist. There are two different conformations of the peptide linkage present in protein, these are a-helix and b-conformation. The a-helix always has a right handed arrangement. In b-conformation all peptide chains are streched out to nearly maximum extension and then laid side by side and held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as b-pleated sheet. (c) Since a - D - (+) - glucose and b - D - (+) glucose differ in configuration at C – 1 atom so they are anomers. NOTE Anomers are those diastereomers

that differ in configuration at C – 1 atom. i.e., (c) in the correct answer. 12.

(c)

NOTE Glucose is considered as a typical

carbohydrate which contains –CHO and –OH group.

13. (a) Biuret test produces violet colour on addition of dilute CaSO 4 to alkaline solution of a compound containing peptide linkage. Polypeptides, proteins and urea have - C- NH - (peptide) linkage while || O

carbohydrates have glycosidic llinkages. So, test of carbohydtrates should be different from that of other three. 14. (b) Molisch's Test : This is a general test for carbohydrates. One or two drops of alcoholic solution of a-naphthol is added to 2 ml glucose solution. 1 ml of conc. H2SO4 solution is added carefully along the sides of the test-tube. The formation of a violet ring at the junction of two liquids confirms the presence of a carbohydrate or sugar. 15. (c) With the exception of glycine all the 19 other common amino acids have a uniquely different functional group on the central tetrahedral alpha carbon. H | H — C — COOH | NH 2

glycine

EBD_7764 C-142

16. 17. 18.

19.

Chemistry

(a) 6CO2 + 12NADPH + 18ATP ® C6H12O6 + 12NADP + 18ADP (a) DNA contains ATGC bases So quinoline is not present in DNA. (c) Water-soluble vitamins dissolve in water and are not stored by the body. The water soluble vitamins include the vitamin Bcomplex group and vitamin C. (a) Among 20 naturally occuring amino acids "Cysteine" has '– SH' or thiol functional group. Þ General formula of amino acid ® R–CH—COOH—NH2 Þ Value of R = –CH2–SH in Cysteine.

20. (a) HOCH2 O

HOCH2 O CH2OH CH2OH Aq.KOH O ¾¾¾¾¾ ® – CH3COOK HO HO – O O – C – CH3 OH OH OH Hemiketal Ring opening

HOCH2 OH Å ve

silver mirror test

CH2OH O OH

Tollen 's ¬¾¾¾ ¾ Reagent

OH

(a Reducing sugar) a-hydroxy ketone

C-143

Polymers

29

Polymers 1.

2.

3.

4.

5.

Polymer formation from monomers starts by [2002] (a) condensation reaction between monomers (b) coordinate reaction between monomers (c) conversion of monomer to monomer ions by protons (d) hydrolysis of monomers. Nylon threads are made of [2003] (a) polyester polymer (b) polyamide polymer (c) polyethylene polymer (d) polyvinyl polymer Which of the following is fully fluorinated polymer? [2005] (a) PVC (b) Thiokol (c) Teflon (d) Neoprene Bakelite is obtained from phenol by reacting with [2008] (a) (CH2OH)2 (b) CH3CHO (c) CH3 COCH3 (d) HCHO Buna-N synthetic rubber is a copolymer of : [2009] (a) H2C = CH – CH = CH2 and H5C6 – CH = CH2 (b) H2C = CH – CN and H2C = CH – CHCH2 (c) H2C = CH – CN and H 2 C = CH – C = CH 2

7.

8.

9.

11.

(b) Neoprene

(c) Teflon

(d) Acrylonitrile

(a) Polypropene

(b) Polyvinyl chloride

(c) Bakelite

(d) Glyptal

Which of the following statements about low density polythene is FALSE? [2016] (a) Its synthesis requires dioxygen or a peroxide initiator as a catalyst. (b) It is used in the manufacture of buckets, dust-bins etc.

C H3

(c) Its synthesis requires high pressure.

Cl

6.

(a) Dacron

10. Which polymer is used in the manufacture of paints and lacquers ? [2015]

|

(d)

(a) teflon (b) nylon 6, 6 (c) polystyrene (d) natural rubber Thermosetting polymer, Bakelite is formed by the reaction of phenol with : [2011RS] (a) CH3CHO (b) HCHO (c) HCOOH (d) CH3CH2CHO The species which can best serve as an initiator for the cationic polymerization is : [2012] (a) LiAlH4 (b) HNO3 (c) AlCl3 (d) BaLi Which one is classified as a condensation polymer? [2014]

|

(d) It is a poor conductor of electricity.

H 2 C = CH – C = CH 2 and

H 2C = CH – CH = CH 2 The polymer containing strong intermolecular forces e.g. hydrogen bonding, is [2010]

12. The formation of which of the following polymers involves hydrolysis reaction?[2017] (a) Nylon 6 (b) Bakelite (c) Nylon 6, 6 (d) Terylene

Answer Key 1 (a)

2 (b)

3 (c)

4 (d)

5 (b)

6 (b)

7 (b)

8 (c)

9 (a)

10 (d)

11 (b)

12 (a)

EBD_7764 C-144

1.

2. 3. 4.

Chemistry

(a) Polymerisation starts either by condensation or addition reactions between monomers. Condensation polymers are formed by the combination of monomers with the elimination of simple molecules.Where as the addition polymers are formed by the addition together of the molecules of the monomer or monomers to form a large molecule without elimination of any thing. (b) Nylon is a polyamide polymer. (c) Teflon is polymer of CF2 = CF2. (d) Bakelite is formed by the reaction of formaldehyde (HCHO) and phenol so the correct answer is (d).

5.

6.

OH n

(b) Buna – N is a copolymer of butadiene (CH2 = CH – CH = CH2) and acrylonitrile (CH2 = CHCN). (b) Nylon 6, 6 has amide linkage capable of forming hydrogen bonding.

OH

+ n HCHO

–

O OH HO Adipic acid

+

H2N

NH2

O

Hexamethylenediamine

H2O H

O

O

N

H2N

NH2 + HO Hexamethylenediamine

OH NH2 + HO

O

Adipic acid

H2O

O

H2O Polymerization

H

O N H

7. 8.

H

O

N

N O Nylon

N

H

(b) (c) Lewis acids are the most common compounds used for initiation of cationic polymerisation. The more popular Lewis acids are SnCl4, AlCl3, BF3 and TiCl4.

C-145

Polymers

9.

(a) Except Dacron all are additive polymers. Terephthalic acid condenses with ethylene glycol to give Dacron.

12. (a) Formation of Nylon-6 involves hydrolysis of caprolactum, (its monomer) in initial state.

COOH +

HOOC Terephthalic acid

® HO O – CH2 – CH2 – OH ¾¾ Ethylene glycol

[

CO

dustbins, bottles, pipes etc. Low density polythene is used for insulating electric wires and in the manufacture of flexible pipes, toys, coats, bottles etc.

]

CH 2 - CH 2 - O -

n

O NH Caprolactam

H 2O

H2 N(CH2)5COOH e-Amino Caproic acid

Dacron (Polyester) 10.

(d)

11.

(b)

Glyptal is used in the manufacture of paints and lacquers. High density polythene is used in the manufacture of housewares like buckets,

O Polymerise

O

–(NH(CH2)5–C–NH–(CH2)5–C)–n Nylon-6

EBD_7764

30

Chemistry in Everyday Life 1.

2.

3.

OCOCH3 COOH

The compound

4. is used as [2002]

(a) antiseptic (b) antibiotic (c) analgesic (d) pesticide. Which of the following could act as a propellant for rockets? [2003] (a) Liquid oxygen + liquid argon (b) Liquid hydrogen + liquid oxygen (c) Liquid nitrogen + liquid oxygen (d) Liquid hydrogen + liquid nitrogen Which one of the following types of drugs reduces fever ? [2005] (a) Tranquiliser (b) Antibiotic (c) Antipyretic (d) Analgesic

5.

6.

Aspirin is known as : [2012] (a) Acetyl salicylic acid (b) Phenyl salicylate (c) Acetyl salicylate (d) Methyl salicylic acid Which of the following compounds is not an antacid ? [2015] (a) Phenelzine (b) Ranitidine (c) Aluminium hydroxide (d) Cimetidine Which of the following is an anionic detergent? [2016] (a) Cetyltrimethyl ammonium bromide. (b) Glyceryl oleate. (c) Sodium stearate. (d) Sodium lauryl sulphate.

Answer Key 1 (c)

1. 2.

3.

2 (b)

3 (c)

4 (a)

5 (a)

6 (d)

(c) The given compound is aspirin which is antipyretic and analgesic (b) Liquid hydrogen and liquid oxygen are used as excellent fuel for rockets. H2(l) has low mass and high enthalpy of combustion whereas oxygen is a strong supporter of combustion. (c) An antipyretic is a drug which is responsible for lowering the temperature of the feverish organism to normal but has no effect on normal temperature states. O–COCH3

4.

(a)

COOH Aspirin (Acetyl salicylic acid)

5. 6.

(a) Phenelzine is an antidepressant, while others are antacids. (d) Sodium lauryl sulphate (C11H23CH2OSO– + 3Na ) is an anionic detergent. Glyceryl oleate is a glyceryl ester of oleic acid. Sodium stearate (C17H35COO–Na+) is a soap. Cetyltrimethyl ammonium bromide + é ù ê CH3 (CH 2 )15 N(CH 3 )3 ú Br ë û is a cationic detergent.

C-147

Analytical Chemistry

31

Analytical Chemistry 1.

When H2S is passed through Hg2S we get [2002] (a) HgS (b) HgS + Hg2S (c) Hg2S + Hg

2.

(d) None of these.

5.

How do we differentiate between Fe3+ and Cr3+ in group III? [2002] (a) by taking excess of NH4OH solution (b) by increasing NH4+ ion concentration (c) by decreasing OH– ion concentration (d) both (b) and (c)

3.

Which one of the following statements is correct ? [2003]

6.

(a) From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl (b) Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution (c) On boiling a solution having K+, Ca2+ and

HCO 3- ions we get a precipitate of

4.

K2Ca(CO3)2 (d) Manganese salts give a violet borax bead test in the reducing flame The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is [2004]

Answer Key 1 (c)

2 (b)

3 (a)

4 (a)

5 (c)

6 (b)

(a) Fe4[Fe(CN)6]3

(b) Na3[Fe(CN)6]

(c) Fe(CN)3

(d) Na4[Fe(CN)5NOS]

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is [2010] (a) 59.0 (b) 47.4 (c) 23.7 (d) 29.5 For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M sulphuric acid. The unreacted acid required 10

20 mL of

M sodium hydroxide for complete 10

neutralization. The percentage of nitrogen in the compound is: [2014] (a) 6%

(b) 10%

(c) 3%

(d) 5%

EBD_7764 C-148

1.

Chemistry

(c) When H2S is passed through Hg2S we get a mixture of mercurous sulphide and

5.

mercury (Hg 2S + Hg) . 2.

(c) Moles of HCl taken = 20 × 0.1 × 10– 3 = 2 × 10–3 Moles of HCl neutralised by NaOH solution = 15 × 0.1 × 10–3 = 1.5 × 10–3

(b) When we add NH4Cl, it suppresses the ionisation of NH4OH and prevents the precipitation of higher group hydroxide in gp(III).

Moles of HCl neutralised by ammonia = 2 × 10–3 – 1.5 × 10–3 = 0.5 × 10–3

NOTE Further ferric chloride and

% of nitrogen =

chromium chloride form different colour precipitates with NH4OH. FeCl3 + 3NH4OH ––––– ® Fe(OH)3 ¯ + 3NH4Cl

=

reddish brown

CrCl3 + 3NH4OH ––––– ® Cr(OH)3 + 3NH4Cl (a) Between AgCl and AgI, AgI is less soluble, hence ammonia can dissolve ppt. of AgCl only due to formation of complex as given below: AgCl + 2NH3 ® [Ag (NH3)2]Cl 4.

(a) Prussian blue Fe 4 [Fe(CN) 6 ]3 is formed in lassaigne test for nitrogen. ® 3Na 4 [Fe(CN) 6 + 4Fe3+ ¾¾

Fe4 [Fe(CN)4 ]6 + 12Na + Prussian blue

1.4 ´ 0.5 ´ 10 -3 29.5 ´ 10 -3

´ 100

= 23.7%

Bluish green.

3.

1.4 ´ N ×V ´ 100 w.t. of Substance

6.

(b) % of N =

1.4 ´ meq. of acid mass of organic compound

M meq. of H2SO4 = 60´ ´ 2 = 12 10 M meq. of NaOH = 20´ = 2 10

\ meq. of acid consumed = 12 – 2 = 10 \ % of N =

1.4 ´ 10 = 10% 1.4

Topic-wise Solved Papers Mathematics

1

Sets 1.

A Ç B = A Ç C and A È B = A È C , then [2009] (a) A = C (b) B = C

(c) 2.

(a) 52 (c) 25

If A, B and C are three sets such that

AÇB = f

(d) A = B

Let X ={1,2,3,4,5}. The number of different ordered pairs (Y,Z) that can formed such that

Y Í X , Z Í X and Y Ç Z is empty is : [2012]

3.

(b) 35 (d) 53

æ1 ö If f(x) + 2f ç ç ÷÷ = 3x , x ¹ 0 and èx ø S = {x Î R : f(x) = f(–x)}; then S: [2016] (a) contains exactly two elements. (b) contains more than two elements. (c) is an empty set. (d) contains exactly one element.

Answer Key 1 (b)

1.

2 (b)

3 (a)

(b) Let x Î A and x Î B Û x Î A È B

Û x Î A È C (Q A È B = A È C ) Û x ÎC \ B = C. Let x Î A and x Î B Û x Î A Ç B Û x Î A Ç C (Q A Ç B = A Ç C )

2.

3.

Û x ÎC \B = C (b) Let X = {1,2,3,4,5} Total no. of elements = 5 Each element has 3 options. Either set Y or set Z or none. (Q Y Ç Z = f) So, number of ordered pairs = 35 æ1ö (a) f (x) + 2f ç ÷ = 3x .......(1) èxø 1 3 f ( ) + 2f (x) = .......(2) x x

1 æ1ö ÷=x+ x èxø

Adding (1) and (2) Þ f (x) + f ç Substracting (1) from (2)

æ1ö 3 Þ f (x) - f ç ÷ = - 3x èxø x On adding the above equations

Þ f (x) =

2 -x x

f (x) = f (- x) Þ x2 = 2

2 -2 2 -x= +xÞx = x x x

or x = 2, - 2

EBD_7764

Relations and Functions 1.

Domain of definition of the function 3

f ( x) =

+ log10 ( x 3 - x) , is 4 - x2 ( -1,0) È (1,2) È ( 2, ¥) (b)

2.

2

3.

The graph of the function y = f(x) is symmetrical about the line x = 2, then [2004] (a) f ( x ) = - f (- x ) (b) f (2 + x ) = f (2 - x ) (c) f ( x ) = f ( - x) (d) f ( x + 2) = f ( x - 2)

4.

The domain of the function f ( x) =

[2003]

(a) (a, 2) (c) ( -1,0) È (a,2) (d) (1,2) È (2, ¥) . If f : R ® R satisfies f ( x + y ) = f ( x) + f ( y ) , n

for all x, y Î R and f(1) = 7, then S f (r ) is r=1 [2003] (a) (c)

7 n (n + 1) 2 7 (n + 1) 2

(b) (d)

7n 2

x -x

(b) (– (d) (–

¥ , 0) ¥, ¥)

Answer Key 1 (a)

1.

(a)

2 (a)

3 (b)

f ( x) =

4 (b)

3 4- x

2

Y

+ log10 ( x 3 - x)

-x

x

4 - x 2 ¹ 0; x 3 - x > 0; x ¹ ± 4 and - 1 < x < 0 or 1 < x < ¥ – –1

+

–

+

1

0

\ D = ( -1, 0) È (1, ¥) -

2.

x1

n

n

r =1

1

7 n ( n + 1) 2 (b) Let us consider a graph symm. with respect to line x = 2 as shown in the figure.

x2

X

x=2

From the figure f ( x1 ) = f ( x2 ), where x1 = 2 - x and x2 = 2 + x \ f (2 - x) = f (2 + x)

D = ( -1, 0) È (1, 2) È (2, ¥). (a) f ( x + y ) = f (x ) + f ( y ) . Function should be f (x) = mx f (1) = 7; \ m = 7, f ( x ) = 7 x S f (r ) = 7 S r =

3.

{ 4}

4.

(b)

f ( x) =

1

is

[2011]

(a) (0, ¥ ) (c) (– ¥ , ¥ ) – {0}

7n + (n + 1) .

1

, define if | x | – x > 0 x -x Þ | x | > x, Þ x < 0 Hence domain of f(x) is (– ¥, 0)

Trigonometric Functions 1.

2.

The number of solution of tan x + sec x = 2cos x in [0, 2 p ) is [2002] (a) 2 (b) 3 (c) 0 (d) 1 Let a, b be such that p < a - b < 3p .

6.

21 27 and cos a + cos b = - , 65 65 a -b [2004] then the value of cos 2 3 -6 (b) (a) 65 130

3.

4.

6 65

(d)

-

( a - b) 2

(b)

2 a 2 + b2

(c)

( a + b) 2

(d)

2(a 2 + b2 )

7.

(c) 1

is (a)

(1 - 7 ) 4

(b)

1 , then tan x 2 [2006]

(4 - 7 ) 3

5 4 and sin (a - b) = , 13 5

where 0 £ a , b £

p . Then tan 2a = 4

56 33

(b)

[2010]

19 12

20 25 (d) 7 16 If A = sin2 x + cos4x, then for all real x : [2011]

(c)

8.

(d) 2

If 0 < x < p and cos x + sin x =

Let cos (a + b ) =

(a)

The number of values of x in the interval [0, 3p] satisfying the equation 2sin 2 x + 5 sin x - 3 = 0 is [2006] (a) 4 (b) 6

5.

then : (a) A is false and B is true (b) both A and B are true (c) both A and B are false (d) A is true and B is false

If u = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q then the difference between the maximum and minimum values of u2 is given by [2004] (a)

3 , 2 [2009]

If cos (b – g) + cos (g – a) + cos (a – b) = -

3 130

(1 + 7 ) (4 + 7 ) (d) 4 3 Let A and B denote the statements A : cos a + cos b + cos g = 0 B : sin a + sin b + sin g = 0 (c) –

If sin a + sin b = -

(c)

3

9.

(a)

13 £ A £1 16

(b) 1 £ A £ 2

(c)

3 13 £ A£ 4 16

(d)

3 £ A £1 4

The possible values of q Î( 0, p) such that sin ( q) + sin ( 4q) + sin ( 7q) = 0 are [2011RS]

EBD_7764 M-4

Mathematics

(a)

π 5π π 2π 3π 8π , , , , , 4 12 2 3 4 9

(b)

2p p p 2p 3p 35p , , , , , 9 4 2 3 4 36

(c)

2p p p 2p 3p 8p , , , , , 9 4 2 3 4 9

13. Let f k ( x ) =

11.

[2014]

( p 2 + q 2 ) sin q p cos q + q sin q

(b)

p 2 + q 2 cos q p cos q + q sin q

(c) (d) 12.

1 4

(a)

2π π 4π π 3π 8π , , , , , 9 4 9 2 4 9 The equation esinx – e–sinx – 4 = 0 has : 1 [2012] (a) infinite number of real roots (b) no real roots (c) exactly one real root (d) exactly four real roots ABCD is a trapezium such that AB and CD are parallel and BC ^ CD. If ÐADB = q, BC = p and CD = q, then AB is equal to : [2013]

(a)

)

and k ³ 1. Then f 4 ( x ) - f 6 ( x ) equals

(d)

10.

(

1 sin k x + cosk x where x Î R k

p2 + q2 p 2 cos q + q 2 sin q ( p 2 + q 2 ) sin q ( p cos q + q sin q) 2

tan A cot A + 1 - cot A 1 - tan A can be written as : (a) sinA cosA + 1 (b) secA cosecA + 1 (c) tanA + cotA (d) secA + cosecA

The expression

[2013]

1 12

(b)

1 1 (d) 6 3 14. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30°, 45° and 60° respectively, then the ratio, AB : BC, is : [2015] (a) 1: 3 (b) 2 : 3

(c)

(c) (d) 3 :1 3: 2 15. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is: [2016] (a) 20 (b) 5 (c) 6 (d) 10 16. If 0 £ x < 2p, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0 is: [2016] (a) 7 (b) 9 (c) 3 (d) 5 17. If 5(tan 2x – cos2x) = 2cos 2x + 9, then the value of cos 4x is : [2017] -

(1) (3)

7 9

-

(2)

1 3

(4)

3 5

2 9

Answer Key 1 (b )

2 (d)

16 (a)

17 (a)

3 (a)

4 (a)

5 (c)

6 (b)

7 (a)

8 (d)

9 (d)

10 (b)

11 (a)

12 (b)

13 (b)

14 (c)

15 (b)

M-5

Trigonometric Functions

1.

(b) The given equation is tanx + secx = 2 cos x; Þ sin x + 1 = 2cos2 x

2

æ -21 ö æ -27 ö + cos 2 b + 2 cos a cos b = ç ÷ +ç ÷ è 65 ø è 65 ø

Þ sin x + 1 = 2(1 – sin 2 x); Þ 2sin2x + sin x – 1= 0; Þ (2sin x – 1)(sin x + 1) = 0

Þ 2 + 2 ( cos a cos b + sin a sin b ) =

1 , –1.; 2 Þ x = 30°, 150°, 270°.

2.

(d)

Þ 2 éë1 + cos ( a - b ) ùû =

p < a - b < 3p

æ a - b ö 1170 Þ 4 cos 2 ç ÷= è 2 ø 4425

p a - b 3p a-b < < Þ cos 1

11.

sin A ö and ÷ cos A ÷ cos A ÷ cot A = ø sin A

tan A =

= 1 + sec A cosec A

4±2 5 =2± 5 2

sin x = 2 ± 5 Q t = e sin x Þ e

ö ÷÷ ø

1 1 = [1 - 2sin 2 x cos-2 x][1 - 3sin 2 x cos 2 x] 4 6 =

1 1 1 - = 4 6 12

EBD_7764 M-8

14.

Mathematics (c)

From (1) and (2)

P 15° 15°

A

Q

30°

45° B

C

3a = x + a Þ x = 2a Here, the speed is uniform So, time taken to cover x = 2 (time taken to cover a)

h

\ Time taken to cover a =

60°

Q 16.

PB bisects ÐAPC, therefore

(a)

AB : BC = PA : PC

and in DCPQ, sin60° =

\ 15.

(b)

AB : BC = 2h :

2h 3

=

cos x = 0, cos

17. (a)

3 :1

...(1)

Þ h = 3a

...(2)

h

B

è

2h h Þ PC = 3 PC

h x+a 1 h Þ = Þ 3h = x + a 3 x +a h h tan 60° = Þ 3 = a a

A

ç2 cos Þ 2cos x ç

h Þ PA = 2h PA

tan 30° =

30° x

minutes cos x + cos 2x + cos 3x + cos 4x = 0 Þ 2 cos 2x cos x + 2 cos 3x cos x = 0

æ

Also in DAPQ, sin30° =

60° a

10 minutes = 5 2

5x xö ÷= 0 cos ÷ 2 2ø

5x x = 0 , cos = 0 2 2

p 3p p 3p 7p 9p x =p, , , , , , 2 2 5 5 5 5 We have 5 tan2 x – 5 cos2 x = 2 (2 cos2 x –1 ) + 9 Þ 5 tan2 x – 5 cos2 x = 4 cos2 x –2 + 9 Þ 5 tan2 x = 9 cos2 x + 7 Þ 5 (sec2 x – 1) = 9 cos2 x + 7 Let cos2 x = t 5 Þ - 9t - 12 = 0 t Þ 9t2 + 12t – 5 = 0 Þ 9t2 + 15t – 3t – 5 = 0 Þ (3t – 1) (3t + 5) = 0

Þt=

1 5 as t ¹ – . 3 3

æ1ö 1 cos 2x = 2 cos2 x – 1 = 2 ç ÷ – 1 = – 3 è3ø 2 7 æ 1ö cos 4x = 2 cos2 2x – 1 = 2 ç - ÷ - 1 = 9 è 3ø

Principle of Mathematical Induction 1.

If an = 7 + 7 + 7 + ... ... having n radical signs then by methods of mathematical induction which is true [2002] (a)

an > 7 " n ³ 1

(b)

an < 7 " n ³ 1

(c)

an < 4 " n ³ 1

(d)

an < 3 " n ³ 1

2.

Let

4

S ( K ) = 1 + 3 + 5... + (2 K - 1) = 3 + K 2 .

Then which of the following is true [2004] (a) Principle of mathematical induction can be used to prove the formula (b)

S ( K ) Þ S ( K + 1)

(c)

S (K ) Þ / S ( K + 1)

(d)

S (1) is correct

Answer Key 1 (b)

1.

2 (b)

(b) a1 =

7 < 7. Let am < 7

Then am + 1 =

7 + am Þ a2m + 1 = 7 + am

< 7 + 7 < 14.

Þ am + 1 < 14 < 7; So by the principle of mathematical induction an < 7 " n. \ r = 0,8,16,24,........256 , total 33 values. 2.

(b) S(K) = 1+3+5+...+(2K – 1) = 3 + K2

S (1) :1 = 3 + 1, which is not true Q S (1) is not true. \ P.M.I cannot be applied Let S(K) is true, i.e.

1 + 3 + 5.... + (2 K - 1) = 3 + K 2

Þ 1 + 3 + 5.... + (2 K - 1) + 2 K + 1 = 3 + K 2 + 2 K + 1 = 3 + ( K + 1) 2

\ S ( K ) Þ S ( K + 1)

EBD_7764 M-10

Mathematics

Complex Numbers and Quadratic Equations 1.

2.

3.

4.

5.

6.

z and w are two non zero complex numbers such that | z | = | w| and Arg z + Arg w = p then z equals [2002] (a) w (b) – w (c) w (d) – w If | z – 4 | < | z – 2 |, its solution is given by [2002] (a) Re(z) > 0 (b) Re(z) < 0 (c) Re(z) > 3 (d) Re(z) > 2 The locus of the centre of a circle which touches the circle | z – z1 | = a and | z – z2 | = b externally (z, z1 & z2 are complex numbers) will be [2002] (a) an ellipse (b) a hyperbola (c) a circle (d) none of these If a ¹ b but a2 = 5a – 3 and b2 = 5b – 3 then the equation having a/b and b/a as its roots is [2002] 2 2 (a) 3x – 19x + 3 = 0 (b) 3x + 19x – 3 =0 (c) 3x2 – 19x – 3 = 0 (d) x2 – 5x + 3 = 0. Difference between the corresponding roots of x2+ax+b=0 and x2+bx+a=0 is same and a ¹ b, then [2002] (a) a + b + 4 = 0 (b) a + b – 4 = 0 (c) a – b – 4 = 0 (d) a – b + 4 = 0 Product of real roots of the equation t2 x2 + | x | + 9 = 0 [2002] (a) is always positive (b) is always negative (c) does not exist (d) none of these

7.

8.

5

If p and q are the roots of the equation x2 + px + q = 0, then [2002] (a) p = 1, q = –2 (b) p = 0, q = 1 (c) p = –2, q = 0 (d) p = – 2, q = 1 If z and w are two non-zero complex numbers

p , 2 [2003]

such that zw = 1 and Arg ( z ) - Arg (w ) = then zw is equal to (a) – 1 (c) – i 9.

(b) 1 (d) i

Let Z1 and Z 2 be two roots of the equation Z 2 + aZ + b = 0 , Z being complex. Further ,

assume that the origin, Z1 and Z 2 form an equilateral triangle. Then [2003] (a)

a 2 = 4b

(b)

a2 = b

(c)

a 2 = 2b

(d)

a 2 = 3b

x

æ1+ i ö ÷ = 1 then è1- i ø

[2003]

10. If ç

11.

(a) x = 2n + 1 , where n is any positive integer (b) x = 4n , where n is any positive integer (c) x = 2n , where n is any positive integer (d) x = 4n + 1 , where n is any positive integer.. The value of 'a' for which one root of the quadratic equation (a 2 - 5a + 3) x 2 + (3a - 1) x + 2 = 0 is twice as large as the other is [2003] (a) -

1 3

(b)

2 3

(c) -

2 3

(d)

1 3

M-11

Complex Numbers & Quadratic Equations

12.

The number of real solutions of the equation x 2 - 3 x + 2 = 0 is

13.

14.

[2003]

(a) 3 (b) 2 (c) 4 (d) 1 Let z and w be complex numbers such that z + i w = 0 and arg zw = p. Then arg z equals [2004] 5p p (a) (b) 4 2 3p p (c) (d) 4 4 If

z = x-i y

an d

1 z3

= p + iq,

æ x yö 2 2 çè p + q ÷ø ( p + q ) is equal to

(a) –2 (c) 2

th en [2004]

If | z 2 - 1|=| z |2 +1, then z lies on (a) an ellipse (b) the imaginary axis (c) a circle (d) the real axis

16.

If (1 - p) is a root of quadratic equation

[2004]

x 2 + px + (1 - p) = 0 then its root are [2004]

(a) –1, 2

(b) –1, 1

(c) 0, –1

(d) 0, 1

4, while the equation x 2 + px + q = 0 has equal roots , then the value of ‘q’ is [2004] (a) 4 (b) 12

18.

(d)

If the cube roots of unity are 1, w , w then the

(a) –1, –1 + 2 w , – 1 – 2 w (b) –1, – 1, – 1

p 2

(b) – p

(c) 0

(d) z

20. If w =

-p 2

and | w | = 1, then z lies on

1 z- i 3

(a) an ellipse

[2005] (b) a circle

(c) a straight line

(d) a parabola p æ Pö . If tan ç ÷ and è 2ø 2

æ Qö – tan ç ÷ are the roots of ax 2 + bx + c = 0, a è 2ø ¹ 0 then [2005] (a) a = b + c (b) c = a + b

(c) b = c (d) b = a + c 22. If both the roots of the quadratic equation x 2 - 2 kx + k 2 + k – 5 = 0 are less than 5, then k lies in the interval [2005] (a) (5, 6] (b) (6, ¥ )

23. The value of

(d) [4, 5] 10

æ

å çè sin

k =1

(a) i (c) – 1

2

2

2

2

(a)

[2005]

2k p 2k p ö + i cos ÷ is 11 11 ø [2006] (b) 1 (d) – i

24. If z 2 + z + 1 = 0 , where z is complex number,, then the value of

49 4

roots of the equation ( x –1)3 + 8 = 0, are

arg z1 – arg z2 is equal to

(c) (– ¥ , 4)

If one root of the equation x 2 + px + 12 = 0 is

(c) 3

2 (d) – 1, 1 + 2 w , 1 + 2 w 19. If z1 and z2 are two non- zero complex numbers such that | z1 + z2 | = | z1 | + | z2 | , then

21. In a triangle PQR, Ð R =

(b) –1 (d) 1

15.

17.

(c) – 1, 1 – 2 w , 1 – 2 w

2

[2005]

1ö æ 2 1ö æ 3 1ö æ çè z + ÷ø + çè z + 2 ÷ø + çè z + 3 ÷ø z z z

......... + æç z 6 + è

(a) 18 (c) 6

1ö ÷ z6 ø

2

2

+

is

[2006] (b) 54 (d) 12

EBD_7764 M-12

25.

If the roots of the quadratic equation x 2 + px + q = 0 are tan30° and tan15°,

26.

respectively, then the value of 2 + q – p is [2006]

(a) less than 4ab

(a) 2

(b) 3

(c) 1ess than – 4ab

(c) 0

(d) 1

(d) greater than 4ab

All the values of m for which both roots of the 2

2

equation x - 2mx + m - 1 = 0 are greater than – 2 but less than 4, lie in the interval [2006]

27.

28.

(a)

-2 < m < 0

(b)

(c)

-1 < m < 3

(d) 1 < m < 4

m>3

If | z + 4 | £ 3, then the maximum value of | z + 1 | is [2007] (a) 6

(b) 0

(c) 4

(d) 10

If the difference between the roots of the equation x2 + ax + 1 = 0 is less than 5 , then the set of possible values of a is [2007] (a)

(3, ¥)

(c) (– 3, 3) 29.

(b)

(- ¥, - 3)

(d)

(-3, ¥) .

The conjugate of a complex number is that complex number is (a)

–1 i –1

(b)

1 then i –1 [2008]

1 i +1

–1 1 (d) i +1 i –1 The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is [2009]

(c)

30.

Mathematics 31. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is : [2009]

(a) 1

(b) 4

(c) 3

(d) 2

(b) greater than – 4ab

4 = 2 , then the maximum value of |z| is z equal to : [2009]

32. If z -

(a)

5 +1

(c)

2+ 2

(b) 2 (d)

3 +1

33. The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals [2010] (a) 1 (b) 2 (c) ¥ (d) 0 34. If a and b are the roots of the equation x2 – x + 1 = 0, then a2009 + b2009 = [2010] (a) –1 (b) 1 (c) 2 (d) –2 35. Let a, b be real and z be a complex number. If z2 + az + b = 0 has two distinct roots on the line Re z =1, then it is necessary that : [2011] (a)

b Î (-1, 0)

(b)

b =1

(c)

b Î (1, ¥)

(d)

b Î (0,1)

36. If w( ¹ 1) is a cube root of unity, and

(1 + w )7 = A + Bw. Then (A, B) equals (a) (1, 1) (c) (–1, 1)

[2011]

(b) (1, 0) (d) (0, 1)

37. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3). Rahul made a mistake in writing down coefficient of x to get roots (3,2). The correct roots of equation are : [2011 RS] (a) 6, 1 (b) 4, 3 (c) – 6, – 1 (d) – 4, – 3

M-13

Complex Numbers & Quadratic Equations

38.

(where [x] denotes the greatest integer £ x ) has no integral solution, then all possible values of f ( x ) = ax 2 + bx + c, g ( x) = a1 x 2 + b1 x + c1and p x = f axlie-ing the x .interval: [2014]

Let for a ¹ a1 ¹ 0,

+ bx + c, g ( x) = a1 x + b1 x + c1and p ( x ) = f ( x ) - g ( x ) . If p (x) = 0 only for x

(a)

= -1 and p (– 2) = 2, then the value of p (2) is : [2011 RS] (a) 3 (b) 9 (c) 6 (d) 18

(b)

39.

40.

41.

2 If z ¹ 1 and z is real, then the point

z -1

represented by the complex number z lies : [2012] (a) either on the real axis or on a circle passing through the origin. (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing through the origin. (d) on the imaginary axis. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a,b,c Î R, have a common root, then a : b : c is [2013] (a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2 If z is a complex number of unit modulus and

æ 1+ z ö argument q, then arg ç equals: [2013] è 1 + z ÷ø p –q (b) 2 (d) p – q

(a) –q (c) q 42.

If z is a complex number such that z ³ 2, then the minimum value of z +

1 : 2

[2014]

3 5 but less than 2 2

5 2 (d) lie in the interval (1, 2) If a Î R and the equation

(c) is equal to 43.

- 3 ( x - [ x ]) + 2 ( x - [ x ]) + a 2 = 0 2

(d) 44. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such that z1 - 2z 2 2 - z1 z2 is unimodular and z2 is not unimodular. Then the point z1 lies on a: [2015] (a) circle of radius 2. (b) circle of radius 2. (c) straight line parallel to x-axis (d) straight line parallel to y-axis. 45. Let a and b be the roots of equation x2 – 6x – 2 = 0. If an = an – bn, for n ³ 1, then the value of a10 - 2a 8 is equal to : 2a 9

[2015]

(a) 3 (b) – 3 (c) 6 (d) – 6 46. The sum of all real values of x satisfying the equation (x 2 - 5 x+ 5)x

2

+ 4x - 60

= 1 is :

[2016] (a) 6 (c) 3 47. A value ofqfor which

5 (a) is strictly greater than 2

(b) is strictly greater than

(c)

( -2, -1) ( -¥, -2 ) È ( 2, ¥ ) ( -1, 0 ) È ( 0,1) (1, 2 )

imaginary, is: (a)

æ 3ö sin -1 ç ÷ ç 4 ÷ è ø

(b) 5 (d) – 4 2 + 3i sinq is purely 1 - 2i sinq [2016]

(b)

æ 1 ö sin -1 çç ÷÷ è 3ø

p p (d) 3 6 48. If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1) (x + 2) + ..... + (x +

(c)

n - 1 ) (x + n) = 10n has two consecutive integral solutions, then n is equal to : [2017] (a) 11 (b) 12 (c) 9 (d) 10

EBD_7764 M-14

Mathematics

Answer Key 1 (b) 16 (c) 31 (b) 46 (c)

1.

2 (c) 17 (d) 32 (a) 47 (b)

3 (b) 18 (c) 33 (a) 48 (a)

4 (a) 19 (c) 34 (b)

5 (a) 20 (c) 35 (c)

(b) Let | z | = | w | = r \ z = reiq, w = reif

6 (a) 21 (b) 36 (a)

7 (a) 22 (c) 37 (a)

8 (a) 23 (d) 38 (d)

\ z = rei(p–f) = reip . e–if = –re–if = – w .

2.

3.

4.

a b Thus, the equation having & as its b a roots is

5.

10 (b) 25 (b) 40 (a)

11 (b) 26 (c) 41 (c)

12 (c) 27 (a) 42 (d)

13 (c) 28 (c) 43 (c)

14 (a) 29 (c) 44 (a)

15 (b) 30 (d) 45 (a)

\ a + b = –a, ab = b and g + d = –b, g d = a. Given |a – b| = |g – d| Þ (a – b)2 = (g – d)2 Þ (a+ b)2 – 4ab = (g + d)2 – 4gd 2 Þ a – 4b = b2 – 4a Þ (a2 – b2) + 4(a – b)= 0 Þ a + b + 4 = 0 (Q a ¹ b)

where q + f = p.

[Q w = re–if] (c) Given | z – 4 | < | z – 2 | Let z = x + iy Þ | (x – 4) + iy) | < | (x – 2) + iy | Þ (x – 4)2 + y2 < (x – 2)2 + y2 Þ x2 – 8x + 16 < x2 – 4x + 4 Þ 12 < 4x Þ x > 3 Þ Re(z) > 3 (b) Let the circle be |z – z0| = r. Then according to given conditions |z0 – z1| = r + a and |z0 – z2|= r + b. Eliminating r, we get |z0 – z1| –|z0 – z2| = a – b. \ Locus of centre z0 is |z – z1| –|z – z2| = a – b, which represents a hyperbola. (a) We have a2 = 5a – 3 and b2 = 5b – 3; Þ a & b are roots of equation, x2 = 5x – 3 or x2 – 5x + 3 = 0 \ a + b = 5 and ab = 3

9 (d) 24 (d) 39 (a)

6. 7.

8.

9 (a) Product of real roots = 2 > 0, " t Î R t \ Product of real roots is always positive. (a) p + q = – p and pq = q Þ q (p – 1) = 0 Þ q = 0 or p = 1. If q = 0, then p = 0. i.e.p = q \ p = 1 and q = –2. (a) | z w |=| z || w |=| z || w |=| z w |= 1 Arg( z w) = arg( z ) + arg(w) = - arg( z ) + arg w = -

p 2

\ z w = -1 iq if Let z = r1e and w = r2e , \ z = r1e -iq

æ a b ö ab x2 - x ç + ÷ + =0 è b aø ab

Now | zw |= 1 Þ r1r2 ei( q+f) = 1 Þ r1 r2 =1

æ a 2 + b2 ö Þ x2 - x ç ÷ +1 = 0 è ab ø

Also arg (z) –arg (w) =

or 3x2 – 19x +3 = 0 (a) Let a, b and g, d be the roots of the equations x2 + ax + b = 0 and x2 + bx + a = 0 respectively.

p p Þ q-f = 2 2

Now z w = r1e - iq .r2eif = r1r2e

-i( q-f )

=e

-

ip 2

= –1

M-15

Complex Numbers & Quadratic Equations

9.

(d)

Z 2 + aZ + b = 0

;

13. (c) arg zw = p Þ arg z + arg w = p...(1)

z + iw = 0 Þ z = -iw p \ z = iw Þ arg z = + arg w 2 p Þ arg z = + p - arg z (from (1)) 2 3p \ arg z = 4

Z1 + Z 2 = - a & Z1Z 2 = b 0, Z1, Z 2 form an equilateral D \ 02 + Z12 + Z 2 2 = 0.Z1 + Z1.Z 2 + Z 2 .0 (for an equilateral triangle, Z12 + Z 22 + Z32 = Z1Z 2 + Z 2 Z 3 + Z3 Z1 ) 2 2 Þ Z1 + Z 2 = Z1Z 2

Þ ( Z1 + Z 2 ) 2 = 3Z1Z 2

14. (a)

2

\ a = 3b (b)

= p + iq

Þ z = p3 + (iq )3 + 3 p (iq )( p + iq ) x

10.

1 z3

Þ x - iy = p3 - 3 pq 2 + i (3 p 2 q - q 3 )

x é (1 + i )2 ù æ 1+ i ö ú =1 çè ÷ø = 1 Þ ê 2 1- i êë 1 - i úû

\ x = p3 - 3 pq 2 Þ

x

æ 1 + i 2 + 2i ö x + ç ÷ = 1 Þ (i ) = 1; \ x = 4n ; n Î I 1 1 + è ø

11.

(b) Let the roots of given equation be a and 2a then a + 2a = 3a =

Þ a=

(

1 - 3a

1 - 3a

)

é 1 (1 - 3a)2 ù 2 2 ê ú= 2 \ 2 2 9 ëê (a - 5a + 3) ûú a - 5a + 3 2

(a - 5a + 3)

= 9 or 9a 2 - 6 a + 1

= 9 a 2 - 45a + 27

or 39a = 26 or a = (c)

y = q2 - 3 p2 q

x y + = -2 p 2 - 2q2 p q

æ x yö \ ç + ÷ ( p 2 + q 2 ) = -2 è p qø

a - 5a + 3

3 a 2 - 5a + 3

(1 - 3a) 2

\

2

2 and a.2a = 2a 2 = 2 a - 5a + 3

12.

y = q3 - 3 p 2 q Þ

x = p 2 - 3q2 p

15. (b)

| z 2 - 1|=| z |2 +1 Þ| z 2 - 1|2 = ( zz + 1) 2 Þ ( z 2 - 1)( z 2 - 1) = ( zz + 1) 2 Þ z 2 z 2 - z 2 - z 2 + 1 = z 2 z 2 + 2zz + 1 2 ( z + z )2 = 0 Þ z = - z Þ z 2 + 2 zz + z Þ =0 Þ z is purely imaginary 1 Let z = r (cosq + i sinq)

Then | z 2 - 1|=| r 2 (cos 2q + i sin 2q) - 1 | = r 4 - 2r 2 cos 2q + 1

2 3

2

and

| z 2 - 1|2 = (| z |2 +1) 2 2

x - 3 x + 2 = 0 Þ| x | -3 | x | +2 = 0

Þ r 4 - 2r 2 cos 2q + 1 = r 4 + 2r 2 + 1

( x - 2)( x - 1) = 0

Þ 2 cos 2 q = 0 Þ cos q = ±

x = 1, 2 or x = ±1, ±2

\ z lies on imaginary axis.

\ No.of solution = 4

p 2

EBD_7764 M-16

Mathematics

2 We know that, if z1 + z2 = z1 + z2 then origin, z1 and z2 are collinear

Þ arg ( z1 ) = arg ( z2 )

2 2 As per question z + ( -1) = z + -1

( )

Þ arg z 2 = arg ( -1) p 2

Þ z lies on imaginary axis. (c) Let the second root be a. Then a + (1 - p ) = - p Þ a = -1 Also a.(1 - p) =1 - p Þ (a - 1)(1 - p) = 0 Þ p =1[Q a = -1] \ Roots are a = -1 and p - 1 = 0

17.

(d) 4 is a root of x 2 + px + 12 = 0

Þ 16 + 4 p + 12 = 0 Þ p = -7 Now, the equation x 2 + px + q = 0 has equal roots. \ p 2 - 4q = 0 Þ q =

18.

(c)

æ Pö æ Qö 21. (b) tan ç ÷ , tan ç ÷ are the roots of è 2ø è 2ø ax 2 + bx + c = 0

Þ 2arg ( z ) = p Þ arg ( z ) =

16.

æ 1ö ç 0, ÷ is same hence z lies on bisector è 3ø of the line joining points (0, 0) and (0, 1/ 3). Hence z lies on a straight line.

p 2 49 = 4 4

b æ Pö æ Qö tan ç ÷ + tan ç ÷ = è 2ø è 2ø a æ Pö æ Qö c tan ç ÷ × tan ç ÷ = è 2ø è 2ø a æ Pö æ Qö tan ç ÷ + tan ç ÷ è 2ø è 2ø æ P Qö = tan ç + ÷ = 1 è 2 2ø æ Pö æ Qö 1 - tan ç ÷ tan ç ÷ è 2ø è 2ø

b b a c a =1 Þ Þ - = c a a a 1a Þ – b = a – c or c = a + b. 22. (c) both roots are less than 5 -

Y axis

1/ 3 ( x - 1)3 + 8 = 0 Þ ( x - 1) = (-2) (1)

Þ x – 1 = – 2 or -2w or - 2w 2 19.

z2 are collinear and are to the same side of origin; hence arg z1 – arg z2 = 0. 20.

0

or x = – 1 or 1 – 2 w or 1 – 2 w 2 . (c) | z1 + z2 | = | z1 | + | z2 | Þ z1 and

(c) As given w =

Þ |w|=

z 1 z- i 3

|z| =1 1 |z- i| 3

1 Þ z = z- i 3 Þ distance of z from origin and point

x=5

X axis

then (i) Discriminant ³ 0 (ii) p(5) > 0 Sum of roots 0 ; 25 – 10 k + k2 + k – 5 > 0 or k2 – 9k + 20 > 0 or k (k – 4) –5(k – 4) > 0 or (k – 5) (k – 4) > 0 Þ k Î ( – ¥ , 4 ) U ( – ¥ , 5)

(iii)

(iii)

Þ

b Sum of roots 2k =– = –2 and m + 1 < 4 Þ m > - 1 and m < 3 or,, -1 < m < 3 (a) z lies on or inside the circle with centre (–4, 0) and radius 3 units. Y

ì 10 - 2k p i ü ï ï = i í å e 11 - 1ý ïî k =0 ïþ

Im.

2p 4p é ù - i - i 11 ê = i 1+ e + e 11 + ....11 terms ú - i ê ú ë û

(-7, 0)

(-4, 0)

Real

(-1, 0)

X'

X

( )

é 2 p 11 ù é 1 - e - 2p i ù ê1 - - 11 ú e ú = iê ú-i = iê 2p - i 2p - iú ê ê - i ú 11 ë1 - e û ëê 1 - e 11 ûú

=i×0–i =–i 24.

25.

(d)

(b)

Y'

From the Argand diagram maximum value of | z + 1| is 6 |z+1|=|z+4–3| £ | z + 4 | + | –3 | £ | 3 | + | – 3|

[Q e -2 pi = 1]

z 2 + z + 1 = 0 Þ z = w or w 2 1 2 So, z + = w + w = -1 z 1 z 2 + 2 = w 2 + w = -1, z 1 z 3 + 3 = w3 + w3 = 2 z 1 1 4 z + 4 = -1, z 5 + 5 = -1 z z 1 and z 6 + 6 = 2 z \ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12

x 2 + px + q = 0 Sum of roots = tan30° + tan15° = – p Product of roots = tan30° . tan15° = q

-p tan 30° + tan15° tan 45° = = =1 1 - tan 30°. tan15° 1 - q

Þ – p = 1- q Þ q - p = 1 \ 2+ q - p = 3

Þ | z + 1 | £ 6 Þ | z + 1|max = 6

28. (c) Let a and b are roots of the equation x2 + ax + 1 = 0 a + b = – a and ab = 1 given | a - b | < 5 Þ

29. 30.

(a + b)2 - 4ab < 5

(Q

(a - b) 2 = (a + b ) 2 - 4ab

Þ a 2 - 4 < 5 Þ a2 – 4 < 5 Þ a2 – 9 < 0 Þ a2 < 9 Þ – 3 < a < 3 Þ a Î (–3, 3) (c) æç 1 ö÷ = 1 = –1 è i –1ø –i –1 i + 1 (d) Let the roots of equation x2 – 6x + a = 0 be a and 4 b and that of the equation x2 –cx + 6 = 0 be a and 3 b .Then a + 4b = 6 ; 4a b = a and a + 3b = c ; 3a b = 6 Þ a=8 \ The equation becomes x2 – 6x + 8 = 0

)

EBD_7764 M-18

31.

Þ (x –2) (x – 4) = 0 Þ roots are 2 and 4 Þ a = 2, b = 1 \ Common root is 2. (b) Given that roots of the equation bx2 + cx + a = 0 are imaginary \ c2 – 4ab < 0 ....(i) 2 2 2 Let y = 3b x + 6 bc x + 2c Þ 3b2x2 + 6 bc x + 2c2 – y = 0

Mathematics 34.

(b)

x2 - x + 1 = 0 Þ x =

x=

1± 3 i 2

a=

1 3 +i = -w2 2 2

Þ 36 b2c2 – 12 b2 (2c2 – y ) ³ 0 Þ 12 b2 (3 c2 – 2 c2+ y ) ³ 0

b=

1 i 3 = -w 2 2

Þ c2 + y ³ 0

a = cos

As x is real, D ³ 0

Þ

y ³ – c2 2 But from eqn. (i), c < 4ab or – c2 > – 4ab \ we get y ³ – c2 > – 4ab Þ y > – 4 ab 32.

(a) Given that z N

Þ Þ Þ

4 4 + z -z

z

=

z-

Þ

z

£ 2+

Þ

z

2

w £

4 4 + z z

–2 z -4£ 0

æ 2 + 20 ö çç z – ÷÷ 2 è ø

æ 2 - 20 ö çç z – ÷÷ £ 0 2 è ø

( z – (1 + 5) ) ( z – (1 - 5) ) £ 0 ( - 5 + 1) £ z £ ( 5 + 1)

z max = 5 + 1 Hence equivalence relation. (a) Let z = x + iy

T

is

an

Þ x=0

z - 1 = z - i ( x - 1) 2 + y 2 = x 2 + ( y - 1) 2

Þ x= y z + 1 = z - i ( x + 1) 2 + y 2 = x 2 + ( y - 1) 2

Only (0, 0) will satisfy all conditions. Þ Number of complex number z = 1

Q a2 + 1 ³ 1

ax 2 + bx + c = 0 Now Sachin’s equation

z - 1 = z + 1 ( x - 1) 2 + y 2 = ( x + 1) 2 + y 2

Þ Re z = 0

= -w2 - w = 1 (c) Since both the roots lie in the line Re z = 1 i.e., x = 1, hence real part of both the roots are 1. Let both roots be 1 + ia and 1 – ia Product of the roots, 1 + a2 = b

\b ³ 1 Þ Q b Î (1, ¥) 36. (a) (1 + w)7 = A + Bw (–w2)7 = A + Bw – w2 = A + Bw 1 + w = A + Bw Þ A = 1, B = 1. 37. (a) Let the correct equation be

4 z

Þ

33.

z-

p p p p + i sin , b = cos - i sin 3 3 3 3

a2009 + b2009 = (-w2 )2009 + ( -w)2009

35.

4 =2 z o

1± 1- 4 2

ax 2 + bx + c ' = 0 Roots found by Sachin’s are 4 and 3

Rahul’s equation, ax 2 + b ' x + c = 0 Roots found by Rahul’s are 3 and 2 b - =7 ....(i) a c =6 ...(ii) a From (i) and (ii), roots of the correct equation x 2 - 7 x + 6 = 0 are 6 and 1. 38. (d) p (x) = 0 f ( x ) = g ( x) Þ Þ

ax2 + bx + c = a1x 2 + b1 x + c1

M-19

Complex Numbers & Quadratic Equations

Þ (a - a1 ) x + (b - b1 ) x + (c - c1 ) = 0. It has only one solution, x = – 1 b - b1 = a - a1 + c - c1 ...(1) Þ 2

vertex = ( -1, 0)

æ 1+ z ö æ 1+ z ö \ arg ç = arg = arg (z) = q. ç 1÷ è 1 + z ÷ø çè 1 + ÷ø z

42.

(d) We know minimum value of |Z1 + Z2| is | |Z1|

b - b1 = -1 2 (a - a1 )

Þ

– |Z2|| Thus minimum value of Z +

Þ Þ b - b1 = 2 ( a - a1 ) ....... 2

...(2)

|Z |-

Now p (– 2) = 2 Þ f (– 2) – g (– 2) = 2 Þ 4a – 2b + c – 4a1 + 2b1 – c1 = 2 Þ 4 (a – a1) – 2 (b – b1) + (c – c1) = 2 ...(3) From equations, (1), (2) and (3) a - a1 = c - c1 =

1 ( b - b1 ) = 2 2

Now, p ( 2) = f ( 2) - g (2)

= 4 ( a - a1 ) + 2 ( b - b1 ) + ( c - c1 ) 39.

= 8 + 8 + 2 = 18 (a) Since we know z = z if z is real.

2-

1 1 1 < Z + < 2+ 2 2 2

3 1 5 < Z+ < 2 2 2 43. (c) Consider –3(x – [x])2 + 2 [x – [x]) + a2 = 0 Þ 3{x}2 – 2{x} –a2 = 0 (Q x – [x] = {x}) Þ

2 æ ö 3 ç {x}2 - {x} ÷ = a2 , a ¹ 0 3 è ø

Þ

2ö æ a 2 = 3{x} ç {x} - ÷ è 3ø

2

Þ z .z - z 2 = z .z - z 2

1/3

Þ z 2 ( z - z ) - ( z - z )( z + z ) = 0

(

)

–1/3

2 Þ (z - z ) z -(z + z ) = 0

Either z - z = 0 or z 2 - ( z + z ) = 0 Either z = z Þ real axis

40.

41.

1 1 £| Z | + 2 2

Þ

2 2 Þ zzz - z = z. z . z - z

2

£ Z+

Since, | Z |³ 2 therefore

z2 z2 = z-1 z -1

Therefore,

1 2

1 is 2

or z 2 = z + z Þ zz - z - z = 0 represents a circle passing through origin. (a) Given equations are x2 + 2x + 3 = 0 …(i) ax2 + bx + c = 0 …(ii) Roots of equation (i) are imaginary roots. According to the question (ii) will also have both roots same as (i). Thus a b c = = = l (say) 1 2 3 Þ a = l, b = 2l, c = 3l Hence, required ratio is 1 : 2 : 3 (c) Given | z | = 1, arg z = q 1 As we know, z = z

2/3

-2 £ a2 < 1 3 (by graph) Since , x is not an integer \ a Î (-1,1) - {0}

Now, {x} Î (0,1) and

Þ a Î (-1, 0) È (0,1) 44.

(a)

z1 - 2z 2 =1 2 - z1 z2 Þ z1 - 2z 2

2

= 2 - z1z2

2

Þ (z1 - 2z 2 )(z1 - 2z 2 ) = (2 - z1z2 )(2 - z1z2 ) Þ (z1 - 2z 2 )(z1 - 2z2 ) = (2 - z1z2 )(2 - z1z 2 )

EBD_7764 M-20

Mathematics Þ

(z1z1) - 2z1z2 - 2z1z 2 + 4z 2 z2

=

4 - 2z1z 2 - 2z1z2 + z1z1z 2 z2

Þ

z1 + 4 z 2

Þ

z1 + 4 z 2

2

2

= 4 + z1

2

2

– 4 – z1

(z

)(

2

- 4 1 - z2

1

Q

z2 ¹ 1

\

z1

Þ

z1 = 2

2

2

2

2

z2 z2

Case II x2 – 5x + 5 = –1 and x2 + 4x – 60 has to be an even number Þ x = 2, 3 where 3 is rejected because for x = 3, x2 + 4x – 60 is odd. Case III x2 – 5x + 5 can be any real number and x2 + 4x – 60 = 0 Þ x = –10, 6 Þ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3 Rationalizing the given expression

2

2

=0

)=0 47.

(b)

(2 + 3isin q)(1 + 2isin q) 1 + 4sin 2 q

=4

For the given expression to be purely imaginary, real part of the above expression should be equal to zero.

Þ Point z1 lies on circle of radius 2. 45.

Þ

6 ± 36 + 8 (a) a, b = = 3 ± 11 2 a = 3 + 11 , b = 3 - 11

(

\ an = 3 + 11

) ( n

– 3 – 11

)

48.

n

=

(

3 + 11

) ( 10

– 3 – 11

(

)

é 2 ê 3 + 11 ë

( ) ( 9 ) - (3 - 11) ùúû – 2 3 + 11

8

+ 2 3 - 11

)

8

9

( 3 + 11)8 éêë( 3 + 11)2 – 2ùúû + ( 3 - 11)8 éêë2 - (3 - 11)2 ùúû = 9 9ù é 2 ê( 3 + 11 ) - (3 - 11) ú ë û

( 3 + 11)8 ( 9 + 11 + 6 11 – 2) + (3 - 11)8 (2 - 9 - 11 + 6 11) = 9 9ù é 2 ê( 3 + 11 ) – ( 3 - 11) ú ë û

( )9 – 6 (3 - 11)9 6 = = =3 9 9ù é 2 2 ê( 3 + 11) - ( 3 - 11 ) ú ë û 6 3 + 11

46.

(c)

(x 2 - 5 x + 5) x

2

+ 4x - 60

=1

Case I x2 – 5x + 5 = 1 and x2 + 4x – 60 can be any real number Þ x = 1, 4

=0

Þ sin 2 q =

å

(x + r - 1)(x + r) = 10n

å

(x 2 + xr + (r – 1)x + r 2 - r = 10n

r =1 n

10

2

1 + 4 sin q 1 Þ sin q = ± 3 (a) We have n

a10 – 2a 8 2a 9

2 - 6 sin 2 q

1 3

r= 1

n

2 Þ å (x + (2r - 1)x + r(r - 1) = 10n r= 1

Þ nx2 + {1 + 3 + 5 + .... + (2n – 1) }x + {1.2 + 2.3 +.... + (n – 1) n} = 10 n (n - 1) n(n + 1) = 10n Þ nx2 + n2 x + 3 n 2 - 31 Þ x2 + nx + =0 3 Let a and a + 1 be its two solutions (Q it has two consequtive integral solutions) Þ a + (a + 1) = – n -n - 1 Þa= ...(1) 2 n 2 - 31 ...(2) Also a (a+1) = 3 Putting value of (1) in (2), we get 2 æ n + 1ö æ 1 - n ö n - 31 -ç = ÷ ç ÷ è 2 øè 2 ø 3

Þ n2 = 121 Þ n = 11

6

Linear Inequality 1.

2.

If a, b, c are distinct +ve real numbers and a2 + b2 + c2 = 1 then ab + bc + ca is [2002] (a) less than 1 (b) equal to 1 (c) greater than 1 (d) any real no. If x is real, the maximum value of

3x 2 + 9 x + 17 3x 2 + 9 x + 7

is

[2006] 1 4

(a)

(d)

Statement-1 : For every natural number n³ 2,

1 1

+

1 2

+ ......... +

1 n

> n.

Statement-2 :For every natural number n ³ 2,

n(n + 1) < n + 1.

[2008]

(a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false

(b) 41

(c) 1

3.

17 7

Answer Key 1 (a)

1.

2 (b)

3 (b)

(a) Q (a – b)2 + (b – c)2 + (c – a)2 > 0 Þ 2(a2 + b2 + c2 – ab – bc – ca) >0

\ Max value of y is 41

Þ 2 > 2(ab + bc + ca) Þ ab + bc + ca < 1 2.

(b)

y=

3 x 2 + 9 x + 17

Given f ( x) =

3x 2 + 9 x + 17 3 x2 + 9 x + 7

3 x2 + 9 x + 7

2

3x ( y - 1) + 9 x ( y - 1) + 7 y - 17 = 0 D ³ 0 Q x is real 81( y - 1)2 - 4 ´ 3( y - 1)(7 y - 17) ³ 0

Þ ( y - 1)( y - 41) £ 0 Þ 1 £ y £ 41

Þ f ( x) = 1 +

10 2

3x + 9 x + 7

Clearly f(x) is maximum when g(x) = 3x2 + 9x + 7 is min.

27 9ö æ Here g(x) = 3 ç x 2 + 3x + ÷ + 7 è ø 4 4

EBD_7764 M-22

Mathematics 2 1 3 = 3 çæ x + ÷ö + è 4 2ø

which is minimum when x = \ f max = 1 +

10 3´

9 3 -9´ +7 4 2

-3 2

=

10 ´ 4 1+ = 41 27 – 54 + 28

3.

(b) Statement 2 is Þ

n(n + 1) < n + 1, n ³ 2

n < n + 1, n ³ 2 which is true

Þ 2 < 3 < 4 < 5 < ------- n

3< nÞ

1 3 1

n£ nÞ

Also

1 1

+

1

2< nÞ

Now

1 1 1 2

>

+

n 1 n 1 3

>

³

2 1 n

>

1 n

;

1 n

\ Adding all, we get

+ ............ +

1 n > = n n n

Hence both the statements are correct and statement 2 is a correct explanation of statement-1.

7

Permutations and Combinations 1.

2.

3.

4.

5.

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are [2002] (a) 216 (b) 375 (c) 400 (d) 720 Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is [2002] (a) 125 (b) 105 (c) 375 (d) 625 Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4, 6 and 7 without repetition. Total number of such numbers are [2002] (a) 312 (b) 3125 (c) 120 (d) 216 The sum of integers from 1 to 100 that are divisible by 2 or 5 is [2002] (a) 3000 (b) 3050 (c) 3600 (d) 3250

7.

9.

(a)

n +1

(c)

n+2

Cr +1

10.

11.

Cr +1

[2003]

(b)

n+ 2

(d)

n +1

Cr

Cr .

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is [2003] (a) 346 (b) 140 (c) 196 (d) 280 The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by [2003] (a) 6! × 5! (b) 6 × 5 (c) 30 (d) 5 × 4

8

(b) 21

C3

(d) 5 38 If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number [2005] (a) 601 (b) 600 (c) 603 (d) 602 At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is [2006] (a) 5040 (b) 6210 (c) 385 (d) 1110 The set S = {1, 2, 3, ......., 12} is to be partitioned into three sets A, B, C of equal size. Thus A È B È C = S, (c)

If Cr denotes the number of combination of n things taken r at a time, then the expression Cr +1 + nC r -1 + 2´n Cr equals

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order [2004] (a) 480 (b) 240 (c) 360 (d) 120 The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [2004] (a)

n

n

6.

8.

12.

A Ç B = B Ç C = A Ç C = f. The number of ways to partition S is [2007] (a) (c)

12! (4!)

(b)

3

12! 3!(4!)

3

(d)

12! (4!) 4 12! 3!(4!) 4

EBD_7764 M-24 13. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? [2008] (a) 8. 6C4. 7C4 (b) 6.7. 8C4 (c) 6. 8. 7C4. (d) 7. 6C4. 8C4 14. From 6 different novels and 3 different dictionaries,4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is: [2009] (a) at least 500 but less than 750 (b) at least 750 but less than 1000 (c) at least 1000 (d) less than 500 15. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is [2010] (a) 36 (b) 66 (c) 108 (d) 3 16. Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3 . Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3. [2011] (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. 17. Statement - 1 : For each natural number n, (n + 1)7–1 is divisible by 7. Statement - 2 : For each natural number

n, n 7 - n is divisible by 7.

[2011 RS]

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1

Mathematics (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true 18. There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points. Then : [2011RS] (a) N £ 100

(b) 100 < N £ 140 (c) 140 < N £ 190 (d) N > 190

{

}

19. If X = 4n - 3n - 1 : n Î N and Y = {9 ( n - 1) : n Î N } , where N is the set of

natural numbers, then X È Y is equal to: [2014] (a) X (b) Y (c) N (d) Y – X (c) 8 (d) 64 20. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least three elements is : [2013, 2015] (a) 275 (b) 510 (c) 219 (d) 256 21. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is : [2015] (a) 120 (b) 72 (c) 216

(d) 192

22. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is : [2016] (a) 52nd (b) 58th (c) 46th (d) 59th 23. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is : [2017] (a) 484 (b) 485 (c) 468 (d) 469

M-25

Permutations and Combinations

Answer Key 1 (d) 14 (c)

1. 2. 3.

4.

5.

2 (c) 15 (c)

3 (d) 16 (a)

4 (b) 17 (a)

5 (c) 18 (a)

6 (c) 19 (b)

(d) Required number of numbers = 5 ´ 6 ´ 6 ´ 4 = 36 ´ 20 = 720. (c) Required number of numbers = 3 ´ 5 ´ 5 ´ 5 = 375 (d) We know that a number is divisible by 3 only when the sum of the digits is divisible by 3. The given digits are 0, 1, 2, 3, 4, 5. Here the possible number of combinations of 5 digitsout of 6 are5C4 = 5, which are as follows– 1 + 2 + 3 + 4 + 5 = 15 = 3 × 5 0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3) 0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3) 0 + 1 + 2 + 4 + 5 = 12 = 3 × 4 0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3) 0 + 1 + 2 + 3 + 4 = 10 ( not divisible by 3) Thus the number should contain the digits 1, 2, 3, 4, 5 or the digits 0, 1, 2, 4, 5. Taking 1, 2, 3, 4, 5, the 5 digit numbers are = 5! = 120 Taking 0, 1, 2, 4, 5, the 5 digit numbers are = 5! – 4! = 96 \ Total number of numbers = 120 + 96 = 216 (b) Required sum = (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100) – (10 + 20 + ... + 100) = 2550 + 1050 – 530 = 3050. (c) n Cr +1 + n Cr -1 + 2 n Cr n

n

n

n

= Cr -1 + Cr + Cr + Cr +1 = n +1Cr + n +1Cr +1 = n + 2 Cr +1

6.

(c) As for given question two cases are possible.

7 (a) 20 (c)

8 (c) 21 (d)

9 (b) 22 (b)

10 (a) 23 (b)

11 (c)

12 (a)

13 (d)

(i) Selecting 4 out of first five question and 6 out of remaining question = 5C4 ´8 C6 = 140 choices. (ii) Selecting 5 out of first five question and 5 out of remaining 8 questions = 5C5 ´8 C5 = 56 choices.

7.

Therefore, total number of choices =140 + 56 = 196. (a) No. of ways in which 6 men can be arranged at a round table = (6 - 1)! = 5! Now women can be arranged in 6 P5

8.

= 6! Ways. Total Number of ways = 6! × 5! (c) Total number of arrangements of letters in the word GARDEN = 6 ! = 720 there are two vowels A and E, in half of the arrangements A preceeds E and other half A follows E. So, vowels in alphabetical order in 1 ´ 720 = 360 2

9.

(b) We know that the number of ways of distributing n identical items among r persons, when each one of them receives at least one item is

n -1

Cr -1

\ The required number of ways

7! 7´ 6 = = 21 2!5! 2 ´1 10. (a) Alphabetical order is A, C, H, I, N, S No. of words starting with A – 5! No. of words starting with C – 5!

=

8 -1

C3-1 = 7C2 =

EBD_7764 M-26

Mathematics

No. of words starting with H – 5! No. of words starting with I – 5! No. of words starting with N – 5! SACHIN – 1 \ sachin appears at serial no 601 10

10

C1 + C2 +

10

= 10 -1 C4 -1 = 9 C3 17. (a) Statement 2 : P ( n ) : n 7 - n is divisible by 7

Put n = 1, 1 – 1 = 0 is divisible by7, which is true Let n = k, P (k) : k7 – k is divisible by 7, true Put n = k + 1

10

C3 + C4

11.

(c)

12.

= 10 + 45 + 120 + 210 = 385 (a) Set S = {1, 2, 3, ...... 12}

7 \ P (k + 1) : (k + 1) – ( k + 1) is div. by 7 P(k + 1) : k7 + 7C1k6 + 7C2k2 +......+ 7C6k + 1 – k – 1, is div. by 7. P(k + 1) : (k7 – k) + (7C1k6 + 7C2k5 +........+ 7C k) is div. by 7. 6 Since 7 is coprime with 1, 2, 3, 4, 5, 6.

AÈ B È C = S, AÇ B = B ÇC = A ÇC = f \ The number of ways to partition = 12C4 × 8C4 × 4C4 =

13.

12! 12! 8! 4! = ´ ´ 4!8! 4!4! 4!0! (4!)3

So 7 C1 , 7 C 2 , ......7 C 6are all divisible by 7

(d) First let us arrange M, I, I, I, I, P, P

\ P(k + 1) is divisible by 7 Hence P(n) : n7 – n is divisible by 7

7! ways 4!2! Ö MÖ IÖIÖ IÖIÖ PÖPÖ Now 4 S can be kept at any of the ticked places in 8C4 ways so that no two S are adjacent. Total required ways

Which can be done in

Statement 1 : n7 - n is divisible by 7

= 6 C4 ´3C1 ´ 4! = 1080

(c) Total number of ways = 3 C2 ´ 9C2

16.

9´8 = 3 ´ 36 = 108 2 (a) The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box empty is same as the number of ways of selecting (r – 1) places out of

=3×

Hence require number of ways

( n + 1) 7 - n7 - 1 + ( n7 - n) ( n + 1) 7 - n7 - 1 is divisible by 7

Hence both Statements 1 and 2 are correct and Statement 2 is the correct explanation of Statement -1. 18. (a) Number of required triangles = 10 C3 -6 C3 10 ´ 9 ´ 8 6 ´ 5 ´ 4 = 120 - 20 = 100 6 6 19. (b) 4n – 3n – 1 = (1 + 3)n – 3n –1 = [nC0 + nC1.3 + nC2.32 +......+ nCn3n] – 3n – 1 = 9 [nC2 +nC3.3+....+nCn.3n–2] \ 4n – 3n – 1 is a multiple of 9 for all n. \ X = {x : x is a multiple of 9} Also, Y = {9 (n – 1) : n ÎN} = {All multiples of 9} Clearly X Ì Y. \ X È Y = Y =

15.

(n – 1) different places, that is

Þ

Þ

out of 3 can be selected in 6 C4 ´3C1 ways Then 4 novels with one dictionary in the middle can be arranged in 4! ways. \ Total ways of arrangement

( n + 1) 7 - ( n + 1) is divisible by 7

is divisible by 7

7! 8 7! 8 C4 = C4 = 7 ´ 6C4 ´ 8C4 4!2! 4!2! (c) 4 novels, out of 6 novels and 1 dictionary

=

14.

Þ

n -1

Cr -1 .

20.

(c) Given n(A) = 2, n(B) = 4, n(A × B) = 8

M-27

Permutations and Combinations

Required number of subsets = 8C + 8C +.... + 8C = 28 – 8C – 8C – 8C 3 4 8 0 1 2 = 256 – 1 – 8 – 28 = 219 21.

(d) Four digits number can be arranged in 3 × 4! ways. Five digits number can be arranged in 5! ways.

22.

Number of integers = 3 × 4! + 5! = 192. (b) ALLMS No. of words starting with _____ A: A

4! = 12 2!

_ _ _ _ _ 4! = 24 L : L _____ M: M

4! = 12 2!

_A _____ S : S

3! =3 2!

: S_ L_ _ _ _ 3! = 6

SMALL ® 58 th word

4 ladies 23. (b)

X

3 ladies Y

3 men

4 men

Possible cases for X are (1) 3 ladies, 0 man (2) 2 ladies, 1 man (3) 1 lady, 2 men (4) 0 ladies, 3 men Possible cases for Y are (1) 0 ladies, 3 men (2) 1 lady, 2 men (3) 2 ladies, 1 man (4) 3 ladies, 0 man No. of ways = 4C3 . 4C3 + (4C2 . 3C1)2 + (4C1 . 3C2)2 + (3C3)2 = 16 + 324 + 144 + 1 = 485

EBD_7764 M-28

Mathematics

8

Binomial Theorem 1.

2.

3.

4.

5.

6.

7.

The coefficients of xp and xq in the expansion of (1+ x )p+q are [2002] (a) equal (b) equal with opposite signs (c) reciprocals of each other (d) none of these If the sum of the coefficients in the expansion of (a + b)n is 4096, then the greatest coefficient in the expansion is [2002] (a) 1594 (b) 792 (c) 924 (d) 2924 The positive integer just greater than (1 + 0.0001)10000 is [2002] (a) 4 (b) 5 (c) 2 (d) 3 r and n are positive integers r > 1, n > 2 and coefficient of (r+2)th term and 3rth term in the expansion of (1 + x)2n are equal, then n equals [2002] (a) 3r (b) 3r + 1 (c) 2r (d) 2r + 1 If x is positive, the first negative term in the expansion of (1 + x)27 5 is [2003] (a) 6th term (b) 7th term (c) 5th term (d) 8th term The number of integral terms in the expansion of ( 3 + 8 5 )256 is [2003] (a) 35 (b) 32 (c) 33 (d) 34 The coefficient of the middle term in the binomial

expansion in powers of x of (1 + ax )4 and of is the same if a equals [2004] 3 10 (a) (b) 5 3 -3 -5 (c) (d) 10 3 8. The coefficient of x n in expansion of (1 + x )(1 - x )n is

[2004]

9.

(a)

( -1)n -1 n

(b)

( -1)n (1 - n)

(c)

( -1)n -1 (n - 1) 2

(d)

(n - 1)

The value of

50

C4 +

6

å 56 - r C3

is

[2005]

r =1

(a)

55

C4

(b)

55

C3

(c)

56

C3

(d)

56

C4 11

7 é ù 10. If the coefficient of x in ê ax 2 + æç 1 ö÷ ú è ø bx ë û

11

-7 é 1 öù equals the coefficient of x in ê ax - æç ú , è bx 2 ÷ø û ë

then a and b satisfy the relation [2005] (a) a – b = 1 (b) a + b = 1 (c) 11.

a =1 b

(d) ab = 1

3 If x is so small that x and higher powers of x

(1 +

3 x) 2

may be neglected, then

æ 1 ö - ç 1 + x÷ è 2 ø

3

1 (1 - x ) 2

may be approximated as

[2005]

(a)

3 1 - x2 8

(b)

3 3x + x2 8

(c)

3 - x2 8

(d)

x 3 2 - x 2 8

12. For natural numbers m, n if (1 - y ) m (1 + y ) n = 1 + a1 y + a2 y 2 + ....... and a1 = a2 = 10, then (m, n) is

[2006]

Binomial Theorem

13.

(a) (20, 45) (b) (35, 20) (c) (45, 35) (d) (35, 45) In the binomial expansion of (a – b)n, n ³ 5, the sum of 5th and 6th terms is zero, then a/b equals [2007] (a)

n -5 6

(b)

5 n-4 The sum of the series

(c)

14.

20

C0 -

20

is (a) 0 (c)

20

C1 +

C2 -

- 20 C10

(d)

20

n-4 5 6 . n -5 [2007]

C3 + .....-..... +

20

Statement -1 :

17.

20.

(d)

20

x +1 x -1 ö æ [2013] çè 2 / 3 1/3 ÷ is x - x + 1 x - x1/ 2 ø (a) 4 (b) 120 (c) 210 (d) 310 21. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 – Tn = 10, then the value of n is : [2013] (a) 7 (b) 5 (c) 10 (d) 8 22. The sum of coefficients of integral power of x in

C10

n

å (r + 1) nCr x r

r =0

[2008] = (1 + x) + nx(1 + x ) . (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false The remainder left out when 82n – (62)2n+1 is divided by 9 is: [2009] (a) 2 (b) 7 (c) 8 (d) 0 Let S1 =

å j ( j - 1) j =1 10

and S3 = å j 2 10C j . j =1

2n

C10

r =0

10

)

1 2

n –1

10

(

2n

20

Statement-2:

16.

)

3 +1 - 3 -1 is : [2012] (a) an irrational number (b) an odd positive integer (c) an even positive integer (d) a rational number other than positive integers The term independent of x in expansion of

(b)

å (r + 1) nCr = (n + 2)2n –1.

n

(

C10

n

15.

M-29 (c) Statement -1 is false, Statement -2 is true . (d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1. 18. The coefficient of x7 in the expansion of (1– x – x2 + x3 )6 is [2011] (a) –132 (b) –144 (c) 132 (d) 144 19. If n is a positive integer , then

C J , S2 =

10

å

j =1

j10C j

[2010]

Statement -1 : S3 = 55 × 29. Statement - 2: S1 = 90 × 28 and S2 = 10 × 28 . (a) Statement -1 is true, Statement -2 is true ; Statement -2 is not a correct explanation or Statement -1. (b) Statement -1 is true, Statement -2 is false.

10

(

the binomial expansion 1 - 2 x

)

50

is :

[2015] 1 50 1 50 3 -1 2 +1 (b) (a) 2 2 1 50 1 50 3 +1 3 (c) (d) 2 2 23. If the number of terms in the expansion of

(

)

(

(

)

( )

)

æ 2 4 ön çç1 - + ÷÷ , x ¹ 0, is 28, then the sum of the è x x2 ø coefficients of all the terms in this expansion, is : [2016] (a) 243 (b) 729 (c) 64 (d) 2187 24. The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + .... + (21C10 – 10C10) is : [2017]

(a) (c)

220 – 210 221 – 210

(b) 221 – 211 (d) 220 – 29

EBD_7764 M-30

Mathematics

A n sw er K ey 1 (a) 16 (a)

1.

2 (c) 17 (b)

3 (d) 18 (b)

5 (d) 20 (c)

6 (c) 21 (b)

7 ( c) 22 ( c)

8 ( b) 23 ( b)

(a) We have tp + 1 = p + qCp xp and tq + 1 = p+qCq xq

2.

4 (c) 19 (a)

p + qC p

p + qC . q

nC r

= [ Remember = nCn – r ] (c) We have 2n = 4096 = 212 Þ n = 12; the greatest coeff = coeff of middle term. So middle term = t7.; t7 = t6 + 1

Þ coeff of t7 = 12C6 =

12! = 924. 6!6!

9 (d) 24 (a)

6.

(d)

7.

4.

5.

= 1+

1 1 + + .......¥ = e < 3 1! 2!

(c) tr + 2 = 2nCr + 1 xr + 1;t3r = 2nC3r – 1 x3r – 1 Given 2nCr + 1 = 2nC3r – 1 ; Þ 2nC2n – (r + 1) = 2nC3r – 1 Þ 2n – r – 1 = 3r – 1 Þ 2n = 4r Þ n = 2r (d) n (n - 1)(n - 2).........( n - r + 1) r Tr +1 = ( x) r! For first negative term, n - r +1 < 0 Þ r > n +1

27 ö æ 32 \ r = 7 . çèQ n = 5 ÷ø 5 Therefore, first negative term is T8 .

Þr >

14 (d)

15 (b)

Cr ( 3)256 - r (8 5)r

Cr (3)

256 - r 2

(5) r / 8

Terms will be integral if 256 - r & r both 2 8 are +ve integer, which is so if r is an integral multiple of 8. As 0 £ r £ 256 (c) The middle term in the expansion of

According to the question

1 æ 1 ö 1 æ 1 ö æ 2ö ç1 - ÷ + ç1 - ÷ + ç1 - ÷ + ... 2! è n ø 3! è n ø è n ø 1 1 1 1 + + + ........+ 1! 2! 3! (9999)!

256

13 (b)

(1 - ax)6 = T4 =6 C3 ( -ax )3 = -20a 3 x3

1 n(n - 1) 1 n( n - 1)(n - 2) 1 = 1 + n. + + + ... 2! n 2 3! n n3

< 1+

Tr +1 = 256

12 (d)

(1 + ax) 4 = T3 =4 C2 (ax )2 = 6a 2 x 2 The middle term in the expansion of

æ 1ö (1 + 0.0001)10000 = ç1 + ÷ , n = 10000 è nø

= 1+ 1+

11 (c)

=

n

3.

(c)

10 (d)

3 10 (b) Coeff. of xn in ( 1+x) (1 – x)n = coeff of xn in 6a 2 = -20a 3 Þ a = -

8.

(1 + x )(1 - n C1 x + n C2 x 2 - .... + (-1)nn Cn x n ) n

n

= ( -1) n C n + ( -1) n -1 C n -1 = ( -1) n + ( -1) n -1 .n

= ( -1)n (1 - n)

Coeff of x n in (1 + x ) (1 - x )

n

= Coeff of xn in

(1 - x) n + Coeff of

x n -1 in (1 - x )

= ( -1)n nCn + ( -1) n -1 nCn-1 = ( -1) n 1 + ( -1)n -1 n

n

M-31

Binomial Theorem

= ( -1) n [1 - n] 9.

(d)

50

C4 +

6

å

Þ

56 - r

r =1

11.

C3

n

We know ë Cr + Cr -1 = 50

= ( C4 +

50

n +1

3

\

Cr ù û = (1 - x )

C3 )

(51 C4 + 51C+352 ) C3 + 53 C3 + 54 C3 + 55 C3

Proceeding in the same way, we get = 10.

(d)

C4 +

C3 =

56

11

C4 .

12. (d) 2 11 - r

11

= Cr (ax )

æ 1ö çè ÷ø bx

=

C5 (a) (b)

11

11 - r

and a2 =

-5

-r

- 2r

m2 + n2 - m - n - 2mn = 10 2

So, n – m = 10 and (m - n) 2 - (m + n) = 20

æ 1 ö = 11Cr (ax 2 )11 - r ç è bx 2 ÷ø r

(1 - y ) m (1 + y ) n

\ a1 = n - m = 10

Again T r + 1 in the expansion 1 ù é ê ax - 2 ú bx û ë

1 3 ù . 2 2 x 2 ú é -3 x 2 ù = -3 x 2 úê ú 8 2! ûë8 û

ì m( m - 1) n(n - 1) ü + - mný y 2 + ..... = 1 + ( n - m) + í 2 2 î þ

...(1)

11

3 1 ö éæ ù . æ 3 x 3.2 x 2 ö ú êç 3 2÷ 2 2 x ÷ - ç1 + + êçè 1 + x + ÷ú ø è 2 2! 2 2! 4 ø úû êë

[1 + n C1 y + n C 2 y 2 + .....]

\ Coefficient of x 7 6

)

3

= [1 - m C1 y + m C2 y 2 - ......]

r

= 11 Cr (a)11 - r (b) - r ( x)22 - 2r - r For the Coefficient of x7, we have 22 – 3r = 7 Þ r = 5 11

-1 2

1 x2

xö ÷ 2ø

(as x3 and higher powers of x can be neglected)

Tr +1 in the expansion é 2 1 ù ê ax + bx ú ë û

(

é ê x = ê1 + + ë 2

=

55

(1 + x) 2 - æçè1 + 1-

+51 C3 +52 C3 + 53 C3 +54 C3 +55 C3

55

C5 (a)6 (b) - 5 = 11C6 a5 ´ (b) - 6

Þ ab = 1. (c) Q x3 and higher powers of x may be neglected

é 55 C3 + 54C3 + 53C3 + 52 C3 ù ú = 50 C4 + ê êë + 51C3 + 50C3 úû

én

11

r

11 - r

( -1) ´ (b ) ( x ) ( x) = Cr (a ) –7 For the Coefficient of x , we have Now 11 – 3r = – 7 Þ 3r = 18 Þ r = 6

\ Coefficient of x - 7

= 11C6 a5 ´ 1 ´ (b) - 6 \ Coefficient of x7 = Coefficient of x–7

13.

Þ m + n = 80 \ m = 35, n = 45 (b) Tr + 1 = (–1)r. nCr (a)n – r. (b)r is an expansion of (a – b)n 5th term = t5 = t4+1 \ = (–1)4. nC4 (a)n–4.(b)4 = nC4 . an–4 . b4 6th term = t6 = t5+1 = (–1)5 nC5 (a)n–5 (b)5 Given t5 + t6 = 0 \ nC4 . an–4 . b4 + (– nC5 . an–5 . b5) = 0 Þ

n! an n! a n b5 . .b4 . =0 4!(n - 4)! a 4 5!(n - 5)! a5

EBD_7764 M-32

Mathematics

+ 2 n +1C2 (63)2 n -1 – ........+ (–1)2n+1 2n+1C2+1 ù û = 63 ×

n !.a n b 4

é 1 b ù =0 4 ê ( n - 4) 5.a ú 4!( n - 5)!.a ë û

Þ

é nC0 (63)n -1 + n C1 (63)n - 2 + n C2 (63)n -3 ë

1 b a n-4 =0 Þ = n - 4 5a b 5 20 20 (d) We know that, (1 + x) = C0 + 20C1x +

or,

14.

15.

20C x2 + ...... 20C x10 + ..... 20C x20 2 10 20 Put x = –1, (0) = 20C0 – 20C1 + 20C2 – 20C3 + ...... + 20C10 – 20C11 .... + 20C20 Þ 0 = 2[20C0 – 20C1 + 20C2 – 20C3 + ..... – 20C9] + 20C10 Þ 20C = 2[20C – 20C + 20C – 20C 10 0 1 2 3 + ...... – 20C9 + 20C10] Þ 20C – 20C + 20C – 20C + .... + 20C 0 1 2 3 10 1 = 20C10 2

(b) We have n

å (r + 1) nCr x r =

r =0

n

n

n

r =0

r =0

+........] + 1 – 63 × é 2n+1C (63) 2n - 2n+1C (63)2n -1 + .......ù + 1 0 1 ë û

= 63 × some integral value + 2 = 82n – (62)2n+1when divided by 9 leaves 2 as the remainder. 17. (b)

j =1

18.

å r. nCr x r + å nCr xr

n

n –1 Cr –1 x r + (1 + x)n =år× r r =1

n

= nx å

r =1

Cr –1 x r –1 +(1 + x )n

r=0

+ ........+

n

Cn-1 (63) + nCn ù û

2n +1 C0 (63) 2n +1 - 2n +1C1 (63)2 n = éë

j =1

3 +1

( 3)

)

2n

2 n -1

-

(

)

3 -1

+ 2 n C3

2n

( 3)

( 3)

2 n -3

2n-5

+ ....ù úû n (Using binomial expansion of (a + b) and (a – b)n) = which is an irrational number.

å (r + 1)n Cr = n × 2n –1 + 2n = (n + 2) × 2n –1.

é nC (63)n + nC (63)n -1 + nC (63)n - 2 1 2 ë 0

10

C j = å 10 9 C j -1

+ 2 n C5

n

16.

(

= 2 é 2n C1 êë

= nx (1+ x) n–1 + (1+ x) n = RHS \ Statement 2 is correct. Putting x = 1, we get

10

= 10 é 9 C0 + 9C1 + 9 C2 + .... + 9 C9 ù = 10.29 ë û (b) (1 – x – x2 + x3)6 = [(1– x) – x2 (1 – x)]6 = (1– x)6 (1 – x2)6 = (1 – 6x + 15x2 – 20x3 + 15x4 – 6x5 + x6) × (1 – 6x2 + 15x4 – 20x6 + 15x8 – 6x10 + x12) Coefficient of x7 = (– 6) (– 20) + (– 20)(15) + (– 6) (–6) = – 144

19. (a) Consider

n –1

\ Statement 1 is also true and statement 2 is a correct explanation for statement 1. (a) (8)2n – (62) 2n + 1 = (64) n – (62)2n + 1 = (63 + 1)n – (63 – 1)2n + 1 =

10

S2 = å j

20.

(c) Given expression can be written as 10

æ 1/ 3 æ x +1 ö ö ç ( x + 1) - çç ÷÷ ç x ÷ø ÷ø è è 10

æ 1 ö = ç x1/ 3 + 1 - 1 ÷ xø è

= (x1/3 – x–1/2)10 General term = Tr+1 = 10Cr (x1/3)10–r(–x–1/2)r

M-33

Binomial Theorem

=

10

Cr

= 10 Cr

10 - r x 3

-r · ( -1) r · x 2

+... +

22.

(1 - 2

10 - r r - = 0 3 2 Þ r=4 So, required term = T5 = 10C4 = 210 (b) We know, Tn = nC3, Tn+1 = n+1C3 ATQ, Tn+1 – Tn = n+1C3 – nC3 = 10 Þ nC2 = 10 Þ n = 5.

(c)

(1 + 2 x )

50

C 4 (2 x ) 4 ...(2)

50

= C0 + C1 2 x - C 2 (2 x )

50

x)

50

+ (1 + 2 x )

= 2 éë 50 C0 + 50C2 22 x +

50

C4 23 x 2 + ...ùû

350 + 1 2 23. (b) Total number of terms = n+2C2 = 28 (n + 2) (n + 1) = 56; x = 6 24. (a) We have (21C1 + 21C2 ...... + 21C10) – (10C1 + 10C2 ..... 10C10) 1 21 [( C1 + .... + 21C10 ) + (21C11 + .... 21C20)] 2 – (210 – 1) (Q 10C1 + 10C2 + .... + 10C10 = 210 – 1)

= 2

...(1) 50

50

Putting x = 1, we get above as

(1 - 2 x )50 =50C0 -50C1 2 x +50C2 ( 2 x )

50

C3 (2 x )3 -

Adding equation (1) and (2)

10- r - r ( -1) r · x 3 2

Term will be independent of x when

21.

50

2

1 21 [2 - 2] - (210 - 1) 2 = (220 – 1) – (210 – 1) = 220 – 210

=

EBD_7764 M-34

Mathematics

9

Sequences and Series 1.

2.

3.

4.

5.

6.

If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equals [2002] (a) log3 4 (b) 1 – log3 4 (c) 1 – log4 3 (d) log4 3 The value of 21/4. 41/8. 81/16 ... ¥ is [2002] (a) 1 (b) 2 (c) 3/2 (d) 4 Fifth term of a GP is 2, then the product of its 9 terms is [2002] (a) 256 (b) 512 (c) 1024 (d) none of these Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is [2002] (a) 5 (b) 3/5 (c) 8/5 (d) 1/5 13 – 23 + 33 – 43 +...+93 = [2002] (a) 425 (b) –425 (c) 475 (d) –475 The sum of the series [2003] 1 1 1 + .......... .. up to ¥ is equal to 1.2 2.3 3.4 æ 4ö èeø

(a) log e ç ÷ 7.

(b) 2 log e 2

(d) log e 2 (c) log e 2 - 1 If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of the a b c squares of their reciprocals, then , and c a b are in [2003] (a) Arithmetic - Geometric Progression (b) Arithmetic Progression (c) Geometric Progression (d) Harmonic Progression.

8.

If Sn =

n

1

å nC

r =0

r

and t n =

n

r

å nC

r =0

r

t , then n Sn

is equal to (a)

[2004]

2n –1 2

(b)

1 n 2 Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some

(c) n – 1

9.

1 n -1 2

(d)

positive integers m, n, m ¹ n, Tm = Tn =

1 , then a – d equals m

(a)

1 1 + m n

1 and n

[2004] (b) 1

1 (d) 0 mn 10. The sum of the first n terms of the series

(c)

12 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + ...

n(n + 1)2 when n is even. When n is odd the 2 sum is [2004]

is

(a)

é n(n + 1) ù ê 2 ú ë û

(c)

n(n + 1)2 4

2

(b)

n2 (n + 1) 2

(d)

3n(n + 1) 2

M-35

Sequences and Series

11.

The sum of series

1 1 1 + + + ..... is 2! 4! 6!

(e2 - 2) e

(a)

[2004]

16. Let a1, a2 , a3 ............ be terms on A.P. If a1 + a2 + ...........a p a1 + a2 + ........... + aq

(e - 1)2 2e

(b)

(e2 - 1) (e2 - 1) (d) 2e 2 Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation [2004]

13.

(a)

x 2 - 18 x - 16 = 0

(b)

x 2 - 18 x + 16 = 0

(c)

x 2 + 18 x - 16 = 0

(d) x 2 + 18 x + 16 = 0 If the coefficients of rth, (r + 1)th, and (r + 2)th m

terms in the the binomial expansion of (1 + y ) are in A.P., then m and r satisfy the equation [2005] m 2 – m (4r – 1) + 4 r – 2 = 0

(b)

m2 – m (4r + 1) + 4 r + 2 = 0

2

(c)

m – m (4r + 1) + 4 r – 2 = 0

(d)

m2 – m (4r – 1) + 4 r 2 + 2 = 0 ¥

å

an , y =

¥

å

bn , z =

¥

å cn

14.

If x =

15.

b, c are in A.P and |a | < 1, | b | < 1, | c | < 1 then x, y, z are in [2005] (a) G. P. (b) A.P. (c) Arithmetic - Geometric Progression (d) H.P. The sum of the series [2005] 1+

(a) (c)

n =0

n=0

n=0

where a,

1 1 1 + + + ....................ad inf. is 4.2! 16.4! 64.6 !

e -1 e e -1 2 e

(b) (d)

e +1 e e +1 2 e

,

p ¹ q , then

a6 a21

[2006] (b)

(c)

2 7

(d)

(a)

n(a1 - an )

(b) (n - 1)(a1 - an )

(c)

na1an

(d)

7 2

11 41 17. If a1, a2, .........., an are in H.P., then the expression a1a2 + a2a3 + .......... + an–1an is equal to [2006]

18. The sum of series infinity is

(n - 1)a1an

1 1 1 - + - ....... upto 2! 3! 4! [2007] 1

1 (b) + e 2 e 2 (c) e–2 (d) e–1 In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals [2007]

(a)

19.

2

2

q

2

41 11

(a)

2

(a)

p2

equals

(c)

12.

=

(a)

(b)

5

(

1 2

(

)

5 -1

)

1 1 (d) 1- 5 5. 2 2 20. The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is [2008] (a) –4 (b) –12 (c) 12 (d) 4 21. The sum to infinite term of the series

(c)

2 6 10 14 ... is + + + + 3 32 33 34 (a) 3 (b) 4 (c) 6 (d) 2 1+

[2009]

EBD_7764 M-36 22. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ... are in an AP with common difference –2, then the time taken by him to count all notes is [2010] (a) 34 minutes (b) 125 minutes (c) 135 minutes (d) 24 minutes 23. A man saves ` 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ` 40 more than the saving of immediately previous month. His total saving from the start of service will be [2011] ` 11040 after (a) 19 months (b) 20 months (c) 21 months (d) 18 months

24.

100

å a2r = a and å a2r –1 = β,

then the common

28.

difference of the A.P. is (a)

a -b

[2011] (b)

b-a

7 (99 - 10-20 ) 9

(c)

7 (179 + 10 -20 ) 81

7 (99 + 10 -20 ) 9 If x, y, z are in A.P. and tan–1x, tan–1y and tan–1z are also in A.P., then [2013] (a) x = y = z (b) 2x = 3y = 6z (c) 6x = 3y = 2z (d) 6x = 4y = 3z Let a and b be the roots of equation px2 + qx + r

34 9

(b)

2 13 9

(c)

61 9

(d)

2 17 9

2

7

9

1

( )

8 30. If (10 ) + 2 (11) 10 + 3 (11) (10 ) + .....

å ( k 3 - ( k - 1)3 ) = n3, for any

natural number n. [2012] (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false. If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then the 150th term of this AP is : [2012] (a) – 150 (b) 150 times its 50th term (c) 150 (d) Zero

1 1 + = 4, then a b [2014]

(a)

(d)

9 9 +10 (11) = k (10 ) , then k is equal to: [2014]

k =1

26.

(b)

the value of | a – b| is:

n

Statement-2:

7 (179 - 10-20 ) 81

= 0, p ¹ 0. If p, q, r are in A.P and

a -b 100

α–β 200 Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000.

(c)

29.

r =1

r =1

(a)

(d)

Let an be the n th term of an A.P. If 100

25.

Mathematics 27. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,....., is [2013]

(a) 100

(b) 110

121 441 (d) 10 100 Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is: [2014]

(c) 31.

(a)

2- 3

(b)

2+ 3

(c) (d) 3 + 2 2+ 3 32. The sum of first 9 terms of the series. 13 13 + 23 13 + 23 + 33 + + + .... 1 1+ 3 1+ 3 + 5 (a) 142 (c) 71

[2015] (b) 192 (d) 96

M-37

Sequences and Series

33. If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between l and n, then G14 + 2G 24 + G 34 equals. [2015] (a) 4 lmn2 (b) 4 l2m2n2 (c) 4 l2 mn (d) 4 lm2n 34. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is : [2016] (a) 1

(b)

7 4

8 4 (d) 5 3 If the sum of the first ten terms of the series

(c) 35.

16 m, then m is equal to : [2016] 5 (a) 100 (b) 99 (c) 102 (d) 101 36. For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then : [2017] (a) a, b and c are in G.P. (b) b, c and a are in G.P. (c) b, c and a are in A.P. (d) a, b and c are in A.P. 37. Let a, b, c Î R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, " x, y Î R,

is

10

then

æ 3 ö2 æ 2 ö2 æ 1 ö2 2 æ 4 ö2 çç1 ÷÷ + çç2 ÷÷ + çç3 ÷÷ + 4 + çç4 ÷÷ + ......., è 5ø è 5ø è 5ø è 5ø

(a) (c)

å f ( n ) is equal to :

[2017]

n =1

255 165

(b) 330 (d) 190

A n sw er K ey 1 (b) 16 (d) 31 (b)

1.

2 (b) 17 (d) 32 (d)

3 (b) 18 (d) 33 (d)

4 (b ) 19 (b ) 34 (d )

5 (a) 20 (b) 35 (d)

6 ( a) 21 ( a) 36 ( c)

7 (d) 22 (a) 37 (b)

8 (d ) 23 (c)

(b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P. Þ 2 log9 (31– x+2) = 1 + log3 (4.3x – 1) Þ log3 (31 – x + 2) = log33 + log3 (4.3x – 1) Þ log3 (31– x + 2) = log3 [3(4 × 3x – 1)] Þ 31– x + 2 = 3 (4.3x – 1) Þ 3.3–x + 2 = 12.3x – 3. Put 3x = t Þ

3 + 2 = 12t - 3 or 12t2 – 5t – 3 = 0; t

1 3 Hence t = - , 3 4

Þ 3x =

3 (as 3 x ¹ - ve ) 4

9 (d) 24 (b)

10 (b) 25 (b)

11 (b) 26 (d)

12 ( b) 27 (c)

13 (c) 28 (a)

14 (d) 29 (b)

15 (d) 30 (a)

æ 3ö è 4ø

Þ x = log3 ç ÷ or x = log3 3 – log3 4 Þ x = 1 – log3 4 2.

(b) The product is P = 21/ 4.22 / 8.23 /16......... = 21 / 4 + 2 / 8 + 3 / 16+..........¥ Now let S = 1 + 2 + 3 + ........¥ 4 8 16 1 1 2 S = + + ........¥ 2 8 16 Subtracting (2) from (1)

Þ

1 1 1 1 S = + + + ........¥ 2 4 8 16

......(1) ......(2)

EBD_7764 M-38

Mathematics

1 1/ 4 1 S= = Þ S =1 2 1 - 1/ 2 2

or

3.

2

4.

a +b =

\ P = 2S = 2 (b) ar4 = 2 3

4

5

6

7

a = 20 1- r Þ a = 20(1 – r)... (i) Also a2 + a2r2 + a2r4 + ... to ¥ =100

Þ

[

8.

]

2

(a)

(d)

a

, ab =

As for given condition, a + b =

1 a

2

n

tn =

n

tn =

n

C0

0 C0 n Cn

+

+

+

1 n

C1

1 n

C1

+

+

n -1 n

Cn-1

1 n

C2

2 n

C2

+

+ .... +

+ .... +

n-2 n

Cn- 2

1 n

Cn

n n

Cn

+ .... +

0 n

C0

t n = \ n Sn 2

(d) Tm = a+ (m – 1) d =

1 n

.....(1)

Tn = a + (n - 1)d =

1 m

.....(2)

(1) – (2) Þ (m - n )d =

æ1 1 ö æ 1 1 ö æ 1 1 ö æ 1 1 ö = ç - ÷ - ç - ÷ + ç - ÷ - ç - ÷..... è1 2 ø è 2 3 ø è 3 4 ø è 4 5 ø é1 1 1 1 ù = 1 - 2 ê - + - ................¥ú 2 3 4 5 ë û æ4ö = 1 - 2[- log(1 + 1) + 1] = 2 log 2 - 1 = logç ÷. -b cè e ø

ax 2 + bx + c = 0, a + b =

1

(d) Sn =

Add,

9.

1 1 1 + ...............¥ 1.2 2.3 3.4 1 1 ö æ1 =ç Tn = ÷ n (n + 1) è n n + 1ø S = T1 - T2 + T3 - T4 + T5 ............¥

7.

c a b Þ , , are in A.P.. a b c

é 1 1 1 ù = += + 2tn (n) ê n .... ú nS n n n C1 Cn úû êë C0

é4´5ù = (45) 2 - 16.ê ú = 2025 – 1600 = 425 ë 2 û

6.

2a c b = + b a c

a b c \ , , & are in H.P.. c a b

=100 1 - r2 Þ a2 = 100(1 – r)(1 + r)... (ii) From (i), a2 = 400(1 – r)2; From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)2 Þ 1 + r = 4 – 4r Þ 5r = 3 Þ r = 3/5. (a) 13 – 23 + 33 – 43 + ........+ 93 = 13 + 23 + 33 +.......+ 93 – 2(23 + 43 + 63 + 83) 2

2c b a2 a - = a c2

On simplification 2a 2 c = ab 2 + bc 2

a2

é 9 ´10 ù =ê - 2.2 3 13 + 2 3 + 33 + 4 3 ú ë 2 û

a 2b 2

b2

a2

Given S ¥ = 20 Þ

5.

a +b

2

8

a ´ ar ´ ar ´ ar ´ ar ´ ar ´ ar ´ ar ´ ar = a9 r36 = (ar4)9 = 29 = 512 (b) Let a = first term of G.P. and r = common ratio of G.P.; Then G.P. is a, ar, ar2

Þ

2

a +

1 b2

1 1 1 - Þd= n m mn

1 Þa-d =0 mn 10. (b) If n is odd, the required sum is

From (1) a =

12 + 2.22 + 32 + 2.42 + ...... + 2.(n - 1) 2 + n2 (n - 1)(n - 1 + 1) 2 + n2 2 [Q (n–1) is even =

M-39

Sequences and Series

\ using given formula for the sum of (n–1) terms.]

a, b, c are in A.P. OR æ 1ö 1 1 2 ç1 - ÷ = 1 - + 1 è yø x y

2 æ n - 1 ö 2 n (n + 1) =ç + 1÷ n = è 2 ø 2

11.

12.

15. (d)

Putting x =

1 1 1 e + e -1 \ + + + ...... = -1 2! 4! 6! 2

1+

e 2 + 1 - 2e (e - 1)2 = = 2e 2e

e2 + e ¥= 2

(b) Let two numbers be a and b then

a+b =9 2

Þ x - 18 x + 16 = 0 m

Cr -1 ,

m

Cr ,

m

Cr +1 are in A.P..

2m Cr = m Cr -1 + m Cr +1 m Cr -1 m Cr +1 Þ 2= + m m Cr Cr

Þ

2

2

m - m(4r + 1) + 4r - 2 = 0 . ¥

(d) x =

a = 1-

1 x

å bn = 1 - b

b = 1-

1 y

¥

1 c = 1z

å an

n=0 ¥

y=

=

1 1- a 1

n=0

1 z = å cn = 1- c n=0

1 e

e+ =

2

=

e +1 2 e

p [2a + ( p - 1)d ] p2 2 1 = 2 q [2a1 + (q - 1)d ] q 2

æ p - 1ö a1 + ç d è 2 ÷ø p = q æ q - 1ö a1 + ç d è 2 ÷ø

For

r m-r + m - r +1 r +1

=

-1 2

2a1 + ( p - 1)d p Þ 2a + (q - 1)d = q 1

2

(c) Given

16. (d)

1 we get 2

1 1 1 + + + ...... 4.2! 16.4! 64.6! 1

x 2 - (a + b ) x + ab = 0

14.

e x + e- x x 2 x 4 x6 = 1+ + + ............. 2 2! 4! 6!

é 1 1 ù \ e + e -1 = 2 ê1 + + + ....ú ë 2! 4! û

and ab = 4 \ Equation with roots a and b is

13.

2 1 1 Þ x, y, z are in H.P.. = + y x z

1 1 1 (b) We know that e = 1 + + + + ....... 1! 2! 3! 1 1 1 -1 and e = 1 - + - + ....... 1! 2! 3!

2b = a + c

a6 a 11 , p = 11, q = 41 Þ 6 = a21 a21 41

1 1 1 1 1 1 =d 17. (d) a - a = a - a =..........= an an -1 2 1 3 2 (say)

Then a1a2 =

a2 - a3 a1 - a2 , , a2 a3 = d d

..........., an -1an = \

an -1 - an d

a1a2 + a2 a3 + ......... + an-1an

EBD_7764 M-40

Mathematics

=

a -a a1 - a2 a2 - a3 + + .... + n -1 n d d d

21. (a) We have

2 6 10 14 S = 1 + + 2 + 3 + 4 + .......¥ 3 3 3 3

1 = [a1 - a2 + a2 - a3 + .... + an -1 - an ] d =

a1 - an d

Also, Þ

1 1 = + (n - 1)d an a1

2 1 4 4 4 S = 1 + + 2 + 3 + 4 + ........¥ 3 3 3 3 3

a1 - an = (n - 1)d a1an

(d) We know that ex = 1 + x +

x 2 x3 + + ........¥ 2! 3!

Put x = – 1 \

19.

1 1 1 e–1 = 1 - 1 + - + ........¥ 2! 3! 4!

1 1 1 1 - + - ........¥ 2! 3! 4! 5! (b) Let the series a, ar, ar2, ..... are in geometric progression. given, a = ar + ar 2 Þ 1= r + r2 Þ r2 + r – 1 = 0 \

e–1 =

Þ r=

-1 ± 1 - 4 ´ -1 2

Þ r=

-1 ± 5 2

5 -1 [Q terms of G.P. are positive 2 \ r should be positive] 20. (b) As per question, a + ar = 12 …(1) ar2 + ar3 = 48 …(2) Þ r=

Þ

1 we get 3

1 1 2 6 10 S = + 2 + 3 + 4 + ........¥ ....(2) 3 3 3 3 3 Subtracting eqn. (2) from eqn. (1) we get

a -a Þ 1 n = (n - 1)a1an d Which is the required result.

18.

Multiplying both sides by

....(1)

ar 2 (1 + r ) 48 = Þ r2 = 4, Þ r = –2 a(1 + r ) 12

(Q terms are = + ve and –ve alternately) Þ a = –12

Þ

2 4 4 4 4 S = + 2 + 3 + 4 + ........¥ 3 3 3 3 3

2 Þ S= 3

4 3 = 4´3 Þ S =3 1 3 2 13

22. (a) Till 10th minute number of counted notes = 1500 n 3000 = [ 2 ´ 148 + ( n - 1)( -2) ] = n [148 - n + 1] 2 n 2 - 149 n + 3000 = 0

Þ n = 125, 24 But n = 125 is not possible \ total time = 24 + 10 = 34 minutes. 23. (c) Let required number of months = n \ 200 × 3 + (240 + 280 + 320 + ... + (n – 3)th term) = 11040 Þ

n-3 [ 2 ´ 240 + (n - 4) ´ 40] 2

= 11040 - 600 Þ (n - 3)[240 + 20n - 80] = 10440

(n - 3)(20n + 160) = 10440 (n - 3)(n + 8) = 522 n2 + 5n - 546 = 0 (n + 26) (n – 21) = 0 \ n = 21 24. (b) Let A.P. be a, a + d , a + 2d ,......... a2 + a4 + ........... + a200 = a Þ Þ Þ Þ

M-41

Sequences and Series

Þ

100 é 2 ( a + d ) + (100 - 1) d ùû = a ....(i) 2 ë

1ö æ 1 ö æ 1 öù éæ 7 êç1 - ÷ + ç1 - 2 ÷ + ç1 - 3 ÷ ú = êè 10 ø è 10 ø è 10 ø ú 9 +¼...up to 20 terms ûú ëê

and a1 + a3 + a5 + ......... + a199 = b 100 Þ [2a + (100 – 1) d ] = β ....(ii) 2 On solving (i) and (ii), we get

20 é 1 æ æ 1ö öù 1 ê ç ç ÷ ÷ú 10 è è 10 ø ø ú 7ê = ê 20 ú 1 9ê ú 110 ê ú ë û

a -b 100 25. (b) nth term of the given series d=

2

= Tn = ( n - 1) + ( n - 1) n + n 2

( n - 1)3 - n 3 ) 3 ( = = n - ( n - 1) 3

=

( n - 1) - n

Þ Sn =

n

å

k =1

é k 3 - ( k - 1) 3 ù Þ 8000 = n3 ë û

Multiply and divide by 9

7 é 9 99 999 ù = ê + + + ...¼up to 20 termsú 9 ë10 100 1000 û

ú úû

7 [179 + (10)–20] 81 (a) Since, x, y, z are in A.P. \ 2y = x + z Also, we have 2 tan–1 y = tan–1x + tan–1 (z)

28.

æ 2y ö -1 æ x + z ö Þ tan–1 ç ÷ = tan çè 1 - xz ÷ø è 1 - y2 ø

Þ 29

x+ z 1- y

2

=

x+z 1 - xz

(Q 2y = x + z)

Þ y2 = xz or x + z = 0 Þ x = y = z (b) Let p, q, r are in AP Þ 2q = p + r ...(i) Given

1 1 + =4 a b

Þ

a +b =4 ab

We have a + b = – q/p and ab =

7 77 777 + + + ...¼+ up to 20 terms 10 100 103

é 1 11 111 ù = 7ê + + +¼... + up to 20 termsú ë10 100 103 û

20 ù

=

Þ n = 20 which is a natural number. Now, put n =1,2,3,....20 T1 = 13 – 03 T2 = 23 – 13

M T20 = 203 – 193 Now, T1 + T2 + --- + T20 = S20 Þ S20 = 203 – 03 = 8000 Hence, both the given statement is true. 26. (d) Let 100th term of an AP is a + (100 –1) d = a + 99d where 'a' is the first term of A.P and 'd' is the common difference of A.P. Similarly, 50th term = a + (50 – 1) d = a + 49d Now, According to the question 100 (a + 99d) = 50 (a + 49d) Þ 2a + 198 d = a + 49d Þ a + 149 d = 0 This is the 150th term of an A.P. Hence, T150 = a + 149 d = 0 27. (c) Given sequence can be written as

7 é179 1 æ 1 ö + ç ÷ ê 9 êë 9 9 è 10 ø

q p = 4 Þ q = - 4r r p

r p

-

Þ

From (i), we have 2( – 4r) = p + r p = –9r

....(ii)

EBD_7764 M-42

Mathematics

q = – 4r r=r

32. (d) nth term of series

Now | a - b | = (a + b ) 2 - 4ab 2

æ -q ö 4r = ç ÷ = è pø p =

30.

2

é n(n + 1) ù êë 2 úû 1 = (n + 1) 2 = 2 4 n

q 2 - 4 pr | p|

1 2 Sum of n term = S (n + 1) 4

16 r 2 + 36 r 2 2 13 = | -9 r | 9

=

(a) Let 109 + 2. (11)(10)8 + 3(11)2 (10)7 + ... + 10(11)9 = k(10)9 Let x = 109 + 2.(11)(10)8 + 3(11)2(10)7 + ... + 10(11)9 Multiplied by

1 é n(n + 1)(2n + 1) 2n(n + 1) ù + + nú 4 êë 6 2 û Sum of 9 terms =

11 on both the sides 10

11 x = 11.108 + 2.(11)2.(10)7 + ...+ 9(11)9 + 10 1110

æ 11ö x ç1 - ÷ = 109 + 11 (10)8 + 112 × (10)7 è 10 ø

33. (d)

+ ... + 119 – 1110

Þ

x = (1110 - 1010 ) - 1110 = -1010 10 Þ x = 1011 = k.109 Given Þ k = 100 (b) Let a, ar, ar2 are in G.P. According to the question a, 2ar, ar2 are in A.P. Þ 2 × 2ar = a + ar2 Þ 4r = 1 + r2 Þ r2 – 4r + 1 = 0

Þ

31.

é æ 11 ö10 ù ê ç ÷ - 1ú x 10 ú - 1110 - = 109 ê è ø ê 11 ú 10 -1 ú ê 10 êë úû

-

4 ± 16 - 4 = 2± 3 2 Since r > 1 r=

\

r = 2 - 3 is rejected

Hence, r = 2 + 3

1é 2 Sn + 2 Sn + n ù û 4ë

=

1 é 9 ´ 10 ´ 19 18 ´ 10 ù + + 9ú 4 êë 6 2 û

=

384 = 96 4

m=

l+n and common ratio of G.P. 2 1

n = r = æç ö÷ 4 èlø

\

G1 = l3/4n1/4, G2 = l1/2n1/2, G3 = l1/4 n3/4 3 2 2 3 G14 + 2G 42 + G34 = l n + 2l n + ln

= ln (l + n)2 = ln × 2m2 = 4lm2n 34. (d) Let the GP be a, ar and ar 2 then a = A + d; ar= A + 4d; ar2 = A + 8d Þ

ar 2 - ar (A + 8d)-(A + 4d) = ar - a (A + 4d)-(A + d)

r=

4 3

35. (d)

æ 8 ö2 æ12 ö2 æ16 ö2 æ 20 ö2 æ 44 ö2 çç 5 ÷÷ + çç 5 ÷÷ + çç 5 ÷÷ + çç 5 ÷÷ ... + çç 5 ÷÷ è ø è ø è ø è ø è ø

S=

16 2 2 2 2 + 3 + 4 + ... + 112 25

(

)

Sequences and Series

=

16 æ11(11 + 1)(22 + 1) ö ç - 1÷ ç ÷ 25 è 6 ø

=

16 16 ´505 = ´101 25 5

Þ

36.

(c)

16 16 m = ´101 5 5

Þ m = 101. We have 9(25a2 + b2) + 25 (c2 – 3ac) = 15b (3 a + c) Þ 225a2 + 9b2 + 25c2 – 75ac = 45ab + 15bc Þ (15a)2 + (3b)2 + (5c)2 – 75ac – 45ab–15 bc= 0

1 [(15 a – 3b)2 + (3b – 5c)2 + (5c – 15a)2 ] = 0 2 it is possible when 15a – 3b = 0, 3b – 5 c = 0 and 5c – 15a = 0 Þ 15a = 3b = 5

Þb=

5c c ,a = 3 3

Þa+b=

c 5c 6c + = 3 3 3

37. (b)

M-43 Þ a + b = 2c Þ b, c, a are in A.P. f(x) = ax2 + bx + c f(1) = a + b + c = 3 Þ f (1) = 3 Now f(x + y) = f(x) + f(y) + xy ...(1) Put x = y = 1 in eqn (1) f(2) = f(1) + f(1) + 1 = 2f(1) + 1 f(2) = 7 Þ f(3) = 12 Now, Sn = 3 + 7 + 12 + ......... tn ...(1) Sn = 3 + 7 + ......... tn–1 + tn ...(2) Subtract (2) from (1) tn = 3 + 4 + 5 + .... upto n terms

tn =

(n 2 + 5n) 2

Sn =

å tn = å

(n 2 + 5n) 2

Sn= 1 é n(n + 1)(2n + 1) 5n(n + 1) ù n(n + 1)(n + 8) + ú= 2 êë 6 2 6 û

S10 =

10 ´11 ´18 = 330 6

EBD_7764 M-44

Mathematics

Straight Lines & Pair of Straight Lines 1.

2.

A triangle with vertices (4, 0), (–1, –1), (3, 5) is [2002] (a) isosceles and right angled (b) isosceles but not right angled (c) right angled but not isosceles (d) neither right angled nor isosceles Locus of mid point of the portion between the axes of x cos a + y sina = p whre p is constant is [2002] (a) x2 + y2 =

(c)

1 x

3.

2

+

1 y

2

4

=

(a) y (cos a + sin a ) + x(cos a - sin a ) = a (b) y (cos a - sin a ) - x(sin a - cos a ) = a (c)

2 p

2

(d)

1 x

2

+

1 y

2

=

4 p2

6.

If the pair of straight lines x 2 - 2 pxy - y 2 = 0

7.

and x 2 - 2qxy - y 2 = 0 be such that each pair bisects the angle between the other pair, then [2003] (a) pq = –1 (b) p = q (c) p = –q (d) pq = 1. Locus of centroid of the triangle whose vertices are (a cos t , a sin t ), (b sin t , - b cos t ) and (1, 0),

of lines ax2 + 2hxy + by2 + 2gx + 2fy +

If the pair c = 0 intersect on the y-axis then

where t is a parameter, is

4.

5.

æ

pö 4ø

the origin makes an angle aç 0 < a < ÷ with the è

positive direction of x-axis. The equation of its diagonal not passing through the origin is [2003]

[2003]

(a) (3x + 1) 2 + (3 y ) 2 = a 2 - b 2

[2002] (a) 2fgh = bg2 + ch2 (b) bg2 ¹ ch2 (c) abc = 2fgh (d) none of these The pair of lines represented by 3ax2 + 5xy + (a2 – 2)y2 = 0 are perpendicular to each other for [2002] (a) two values of a (b) " a (c) for one value of a (d) for no values of a A square of side a lies above the x-axis and has one vertex at the origin. The side passing through

y (cos a + sin a ) + x (sin a - cos a) = a

(d) y (cos a + sin a ) + x (sin a + cos a ) = a .

(b) x2 + y2 = 4p2

p2

10

(b) (3x - 1) 2 + (3 y ) 2 = a 2 - b 2 (c) (3x - 1) 2 + (3 y ) 2 = a 2 + b 2 (d) (3x + 1) 2 + (3 y ) 2 = a 2 + b 2 . 8.

If x1, x2 , x3 and y1, y2 , y3 are both in G.P. with the same common ratio, then the points ( x1, y1 ), ( x 2 , y 2 ) and ( x3 , y3 ) (a) (b) (c) (d)

are vertices of a triangle lie on a straight line lie on an ellipse lie on a circle.

[2003]

M-45

Straight Lines & Pair of Straight Lines

9.

If the equation of the locus of a point

13.

6 x 2 - xy + 4cy 2 = 0 is 3x + 4y = 0, then c

equidistant from the point (a1, b1 ) and

(a2, b2 ) is (a1 - b2 ) x + (a1 - b2 ) y + c = 0 then the value of `c` is

,

[2003] 14.

(a)

a12 + b12 - a 2 2 - b2 2

(b)

1 2 a2 + b2 2 - a12 - b12 2

(c) a12 - a 2 2 + b12 - b2 2 (d) 10.

1 2 ( a1 + a2 2 + b12 + b2 2 ) . 2

of the vertex C is the line

[2004]

(a) 3x - 2 y = 3

(b) 2 x - 3 y = 7

(c) 3x + 2 y = 5

(d) 2 x + 3 y = 9

15.

The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is [2004] (a)

x y x y - = 1 and + =1 2 3 -2 1

x y x y - = -1 and + = -1 (b) 2 3 -2 1

12.

equals [2004] (a) –3 (b) 1 (c) 3 (d) 1 The line parallel to the x- axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0, where (a, b) ¹ (0, 0) is [2005] (a) below the x - axis at a distance of

3 from it 2

(b) below the x - axis at a distance of

2 from it 3

(c) above the x - axis at a distance of

3 from it 2

(d) above the x - axis at a distance of

2 from it 3

Let A(2, - 3) and B ( -2, 3) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2 x + 3 y = 1, then the locus

11.

If one of the lines given by

(c)

x y x y + = 1 and + = 1 2 3 2 1

(d)

x y x y + = -1 and + = -1 2 3 -2 1

16.

If the sum of the slopes of the lines given by x 2 - 2cxy - 7 y 2 = 0 is four times their product

c has the value (a) –2 (c) 2

[2004] (b) –1 (d) 1

17.

If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (–1, 2) and (3, 2) then the centroid of the triangle is [2005] 7ö æ (a) ç - 1, ÷ 3ø è

æ -1 7 ö (b) ç , ÷ è 3 3ø

æ 7ö (c) ç1, ÷ è 3ø

1 7 (d) æç , ö÷ è3 3ø

A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is [2006] (a) x + y = 7

(b) 3 x - 4 y + 7 = 0

(c) 4 x + 3 y = 24

(d) 3 x + 4 y = 25

If (a, a 2 ) falls inside the angle made by the lines y =

x , x > 0 and y = 3x , x > 0 , then a 2

belong to

[2006]

EBD_7764 M-46

18.

19.

(a) æç 0, 1 ö÷ è 2ø

(b) (3, ¥)

æ1 ö (c) ç , 3 ÷ è 2 ø

1ö æ (d) ç - 3, - ÷ 2ø è

Let A (h, k), B(1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1square unit, then the set of values which 'k' can take is given by [2007] (a) {–1, 3} (b) {–3, –2} (c) {1, 3} (d) {0, 2}

(c)

21.

22.

24.

3 x+ y =0 2

(b) x + 3 y = 0

3x + y = 0

3 y = 0. (d) x + 2

1 . Then the 3 circumcentre of the triangle ABC is at the point: [2009]

25.

If one of the lines of my2 + (1– m2) xy – mx2= 0 is a bisector of the angle between the lines xy = 0, then m is [2007] (a) 1 (b) 2 (c) –1/2 (d) –2. The perpendicular bisector of the line segment j o i n i n g P (1, 4) and Q(k, 3) has y-intercept –4. Then a possible value of k is [2008] (a) 1 (b) 2 (c) –2 (d) – 4 The shortest distance between the line y – x = 1 and the curve x = y2 is : [2009] (a)

2 3 8

(b)

3 2 5

(c)

3 4

(d)

3 2 8

Mathematics The lines p(p2 +1)x – y + q = 0 and (p2 + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common line for : [2009] (a) exactly one values of p (b) exactly two values of p (c) more than two values of p (d) no value of p Three distinct points A, B and C are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the

point (–1, 0) is equal to

Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three point. The equation of the bisector of the angle PQR is [2007] (a)

20.

23.

æ5 ö (a) çè , 0÷ø 4

æ5 ö (b) çè , 0÷ø 2

æ5 ö (c) çè , 0÷ø 3

(d) (0, 0)

The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersects L3 at R. [2011] Statement-1: The ratio PR : RQ equals 2 2: 5 Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles. (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

26.

The lines x + y = a and ax – y = 1 intersect each other in the first quadrant. Then the set of all possible values of a in the interval :

M-47 [2013]

Straight Lines & Pair of Straight Lines

[2011RS]

27.

28.

(a)

( 0, ¥ )

(b) [1, ¥)

(c)

( -1, ¥)

(d)

(b)

3y = x –

( -1,1)

(c) y =

3x - 3

(d)

3y = x -1

2 x + 3 y = 9, then the locus of the centroid of the triangle is :

(a) 2 + 2

(b) 2 - 2

(c) 1 + 2

(d) 1 - 2

[2011RS]

(a) x - y = 1

(b) 2 x + 3 y = 1

(c) 2 x + 3 y = 3

(d) 2 x - 3 y = 1

If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3 :2, then k equals : [2012] 29 5

32.

Let PS be the median of the triangle vertices P(2, 2), Q(6, –1) and R(7, 3). The equation of the line passing through (1, –1) and parallel to PS is: [2014] (a) 4x + 7y + 3 = 0 (b) 2x – 9y – 11 = 0 (c) 4x – 7y – 11 = 0 (d) 2x + 9y + 7 = 0 Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d =0 lies in the fourth quadrant and is equidistant from the two axes then [2014] (a) 3bc – 2ad = 0 (b) 3bc + 2ad = 0 (c) 2bc – 3ad = 0 (d) 2bc + 3ad = 0 Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus? [2016]

(b) 5 33. (d)

11 5

A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is : [2012] (a) -

31.

3

The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is : [2013]

(c) 6

1 4

34.

(b) – 4 (d) -

(c) – 2 30.

3

(a) y = x +

If A (2, – 3) and B (– 2, 1) are two vertices of a triangle and third vertex moves on the line

(a)

29.

the reflected ray is

1 2

3 y = 3 gets

æ 1 -8 ö (a) ç , ÷ è3 3 ø

æ -10 -7 ö , ÷ (b) ç è 3 3 ø

reflected upon reaching x-axis, the equation of

(c) (–3, –9)

(d) (– 3, – 8)

A ray of light along x +

A n sw er K ey 1 (a) 16 (c) 31 (b)

2 (d) 17 (c) 32 (d)

3 (a) 18 (a) 33 (a)

4 (a) 19 (c) 34 (a)

5 (a) 20 (a)

6 (a) 21 (d)

7 (c) 22 (d)

8 (b) 23 (a)

9 (b) 24 (a)

10 (d) 25 (b)

11 (a) 26 (b )

12 (c) 27 (b)

13 (a) 28 (c)

14 (a) 29 (c)

15 (c) 30 (b)

EBD_7764 M-48

1.

Mathematics

( 4 + 1) 2 + (0 + 1) 2 = 26 ;

(a) AB =

(3 + 1) 2 + (5 + 1) 2 = 52

BC =

CA = ( 4 - 3) 2 + (0 - 5) 2 = 26 ; In isosceles triangle side AB = CA For right angled triangle, BC2 = AB2 + AC2 So, here BC =

2.

52 or BC2 = 52

4.

- 3 ± 9 + 8 - 3 ± 17 = 2 2 (a) Co-ordinates of A = (a cos a , a sin a ) Equation of OB, Þa =

5.

or ( 26 )2 + ( 26 )2 = 52 So, the given triangle is right angled and also isosceles (d) Equation of AB is x cos a + y sin a = p;

For unique point of intersection f 2 – bc = 0 Þ af 2 – abc = 0. Since abc + 2fgh – af 2 –bg2 – ch2 = 0 Þ 2fgh – bg2 – ch2 = 0 (a) 3a + a2 – 2 = 0 Þ a2 + 3a – 2 = 0.;

C p 4

Y

O

B

O

X

æp ö \ slope of CA = - cot ç + a ÷ è4 ø Equation of CA

x cos a y sin a Þ + = 1; p p x y + =1 p / cos a p / sin a So co-ordinates of A and B are p ö æ p ö æ , 0÷ and ç 0, ; çè è sin a ÷ø cos a ø So coordinates of midpoint of AB are Þ

p ö æ p , çè ÷ = ( x1 , y1 )(let ) ; 2 cos a 2sin a ø p p & y1 = ; 2 cos a 2 sin a Þ cos a = p/2x1 and sin a = p/2y1 ; cos2 a + sin2 a = 1 x1 =

1

4

. + = x2 y 2 p 2 (a) Put x = 0 in the given equation Þ by2 + 2 fy + c = 0. Locus of (x1, y1) is

a

CA ^ r to OB

X

A

1

A

æp ö y = tan ç + a ÷ x è4 ø

M (x1, y1)

3.

B

Y

æp ö y - a sin a = - cot ç + a ÷( x - a cos a ) è4 ø æ æp öö Þ (y - a sin a) ç tan ç + a ÷ ÷ = (a cos a - x) è4 øø è

p æ ö tan + tan a ç ÷ 4 Þ (y - a sin a ) ç ÷ (a cos a - x) p ç 1 - tan tan a ÷ è ø 4 Þ (y - a sin a ) (1 + tan a )

= (a cos a - x)(1 - tan a) Þ (y - a sin a ) (cos a +=sin (aacos ) a - x)(cos a - sin a ) Þ y(cos + sin a ) - a sin a cos a - a sin 2 a 2 = a cos a - a cos a sin a - x(cos a - sin a) Þ y(cos a + sin a ) + x(cos a - sin a ) = a y (sin a + cos a ) + x(cos a - sin a ) = a.

M-49

Straight Lines & Pair of Straight Lines

6.

(a) Equation of bisectors of second pair of straight lines is,

æ h -2 + k ö or çè , ÷ . It lies on 2x + 3y = 1 3 3 ø

qx 2 + 2 xy - qy 2 = 0 ....(1) It must be identical to the first pair x 2 - 2 pxy - y 2 = 0 from (1) and (2)

....(2) 11.

q 2 -q = = Þ pq = -1 . 1 - 2 p -1

7.

(c)

a cos t + b sin t + 1 3 Þ a cos t + b sin t = 3 x - 1 a sin t - b cos t y= 3 Þ a sin t - b cos t = 3 y Squaring and adding,

(1) passes through (4, 3) , Þ

x=

(3 x - 1) 2 + (3 y ) 2 = a 2 + b 2

8.

(b) Taking co-ordinates as æx yö ç , ÷; ( x, y ) & ( xr , yr ) . èr rø Then slope of line joining æ 1ö y ç1 - ÷ è rø y æ x yö = çè , ÷ø , ( x, y ) = r r æ 1ö x x ç1 - ÷ è rø

(b)

( x - a1 ) 2 + ( y - b1 ) 2 2

a 2 - 4 = 0 Þ a = ±2 Þ b = -3 or 1

\ Equations of straight lines are x y x y + = 1 or + =1 2 -3 -2 1 12. (c) Let the lines be y = m1x and y = m2x then 2c 1 m1 + m2 = - and m1m2 = 7 7 Given m1 + m2 = 4 m1m2

= ( x - a2 ) + ( y - b2 )

(a1 - a2 ) x + (b1 - b2 ) y 1 + ( a22 + b22 - a12 - b12 ) = 0 2

1 2 ( a2 + b2 2 - a12 - b12 ) 2 (d) Let the vertex C be (h, k), then the centroid of c=

10.

æ 2 - 2 + h -3 + 1 + k ö , DABC is ç ÷ø è 3 3

2c 4 =- Þc=2 7 7 3 x + 4 y = 0 is one of the lines of the pair Þ

13. (a)

6 x 2 - xy + 4cy 2 = 0 ,

we get 6 x 2 +

3 Put y = - x , 4 2

3 2 æ 3 ö x + 4c ç - x ÷ = 0 è 4 ø 4

3 9c + = 0 Þ c = -3 4 4 (a) The line passing through the intersection of lines ax + 2by = 3b = 0 and Þ 6+

14. 2

4 3 + =1 a b

....(3) Þ 4b + 3a = ab Eliminating b from (2) and (3), we get

and slope of line joining (x, y) and (xr, yr) y ( r - 1) y = = x ( r - 1) x \ m1 = m2 Þ Points lie on the straight line. 9.

2h - 2 + k = 1 Þ 2h + 3k = 9 3 Þ Locus of C is 2x + 3y = 9 x y (a) Let the required line be + = 1 ....(1) a b then a + b = –1 ....(2) Þ

bx - 2ay - 3a = 0 is ax + 2by + 3b + l (bx – 2ay – 3a) = 0

Þ (a + b l ) x + (2b – 2a l )y + 3b – 3 l a = 0 As this line is parallel to x-axis. \ a + b l = 0 Þ l = – a/b a Þ ax + 2by + 3b – (bx – 2ay – 3a) = 0 b 2a 2 3a 2 =0 y+ Þ ax + 2by + 3b – ax + b b

EBD_7764 M-50

Mathematics

17. (c) Clearly for point P,

æ 2a 2 ö 3a 2 y ç 2b + =0 ÷ + 3b + b ø b è

y

y = 3x

æ 2b 2 + 2a 2 ö æ 3b2 + 3a 2 ö yç ÷ = -ç ÷ b b è ø è ø 2

y=

15.

2

-3(a + b ) 2

2

2(b + a )

=

2 • P(a, a )

y=

-3 2

O

So it is 3/2 units below x-axis. (c) Vertex of triangle is (1, 1) and midpoint of sides through this vertex is (– 1, 2) and (3, 2)

(-1, 2)

1

0 Þ 0 \ a+1 0 x+ y = a

G (h, k) B(–2, 1)

.... (1)

ax - y = 1 .... (2) On adding equation (1) and (2), we get x (1 + a ) = 1 + a Þ x = 1 y=a–1 It is in first quadrant so a – 1 ³ 0 Þ a ³1 Þ a Î[1, ¥) Case II : If a < 0 x + y = -a

a = 3h

b - 2 = 3k Third vertex (α, β ) lies on the line

2x + 3 y = 9 2a + 3b = 9

2 ( 3h) + 3 ( 3k + 2) = 9 .... (3)

x (1 + a) = 1 - a x=

2h + 3k = 1

28.

2x + 3 y =1 (c) Let the joining points be A (1,1) and B (2,4). Let point C divides line AB in the ratio 3 : 2. So, by section formula we have æ 3´ 2 + 2´1 3 ´ 4 + 2´1 ö C=ç , ÷ 3+2 ø è 3+2

.... (5)

–1 y = -a -

C (a, b)

b = 3k + 2

ax - y = 1 .... (4) On adding equation (3) and (4), we get 1- a a -1 >0Þ 0 Þ a > –1

.... (6)

1- a -a - a2 - 1 + a >0 = >0 1+ a 1+ a

æ 8 14 ö =ç , ÷ è5 5 ø Since Line 2x + y = k passes through æ 8 14 ö Cç , ÷ è5 5 ø

M-53

Straight Lines & Pair of Straight Lines

29.

\ C satisfies the equation 2x + y = k. 2 + 8 14 Þ + =k Þ k=6 5 5 (c) Equation of a line passing through (x1,y1) having slope m is given by y – y1 = m (x – x1 ) Since the line PQ is passing through (1,2) therefore its equation is (y – 2) = m (x – 1) where m is the slope of the line PQ. Now, point P (x,0) will also satisfy the equation of PQ \ y –2 = m (x –1) Þ 0 – 2 = m (x – 1) -2 Þ – 2 = m (x – 1) Þ x – 1 = m Þ x=

Let Area = f (m) = 2 -

Put f ¢ (m) = 0 Þ m2 = 4 Þ m = ± 2 -4 Now, f '' ( m ) = m3 1 f '' ( m ) m = 2 = - < 0 2 1 f '' ( m ) m = - 2 = > 0 2 Area will be least at m = –2 Hence, slope of PQ is –2. 30.

B

1æ 2ö ç 1 - ÷ ( 2 - m) 2è mø 1 (Q Area of D = ´ base ´ height ) 2

=

1é æ 4 öù 4 - ç m + ÷ú 2 êë è m øû

=2-

3, 0

B' (0, –1) Reflected Ray is

31.

-1 - 0 0- 3

Þ 3y = x - 3 (b) From the figure, we have a = 2, b = 2 2, c = 2 x1 = 0, x2 = 0, x3 = 2

m 2 2 m

Q

(1,2)

O

(0, –1)

A

=

1é 4 ù 2 - m - + 2ú 2 êë m û

(b) Suppose B(0, 1) be any point on given line and co-ordinate of A is ( 3, 0). So, equation of

-2 +1 m Similarly, point Q (0,y) will satisfy equation of PQ \ y –2 = m (x– 1) Þ y – 2 = m (–1) Þ y = 2 – m and OQ = y = 2 – m 1 Area of DPOQ = ( OP )( OQ ) 2 =

=

-1 + 2 m2

Now, f ' ( m ) =

-2 +1 m

Also, OP = ( x - 0 ) 2 + ( 0 - 0 )2 = x

P

m 2 2 m 2

=

y -0 x- 3

EBD_7764 M-54

Mathematics The point of intersection will be

Now, x-co-ordinate of incentre is given as

x -y 1 = = 2 ad - 2bc 4ad - 5bc 8ab - 10ab

ax1 + bx2 + cx3 a+b+c Þ x-coordinate of incentre

= 2- 2

Q

(d) Let P, Q, R, be the vertices of DPQR P (2, 2)

bc - ad 5bc - 4ad = Þ 3bc – 2ad = 0 ab 2ab 34.

Q (6, – 1)

S

Since PS is the median S is mid-point of QR

æ 7 + 6 3 - 1ö æ 13 ö So, S = ç , ÷ = ç ,1÷ è 2 2 ø è2 ø

33.

D

(a)

R (7, 3)

2 -1 2 =Now, slope of PS = 13 9 22 Since, required line is parallel to PS therefore slope of required line = slope of PS Now, eqn of line passing through (1, –1) 2 and having slope - is 9 2 y - (-1) = - ( x - 1) 9 9y + 9 = –2x + 2 Þ 2x + 9y + 7 = 0 (a) Given lines are 4ax + 2ay + c = 0 5bx + 2by + d = 0

x – y +1= 0

C

A

5=0

2+ 2

5bc - 4ad 4 ad - 5bc = -2ab 2 ab Point of intersection is in fourth quadrant so x is positive and y is negative. Also distance from axes is same So x = – y (Q distance from x-axis is –y as y is negative) y=

Þ

O (–1,–2)

7x – y –

2

2(ad - bc) bc - ad = -2ab ab

0

32.

2+ 2+ 2 2

+m=

=

2 ´ 0 + 2 2.0 + 2.2

7x – y

=

x=

Þ

x–y +l =0

B

Let other two sides of rhombus are x–y+l=0 and 7x –y + m = 0 then O is equidistant from AB and DC and from AD and BC

\ -1 + 2 + 1 = -1 + 2 +l Þl= – 3 and -7 + 2 - 5 = -7 + 2 +m Þm= 15

\ Other two sides are x – y – 3 = 0 and 7x – y + 15 = 0 On solving the eqns of sides pairwise, we get the vertices as

æ 1 -8 ö æ -7 -4 ö ç ç , ÷ ÷ , (1, 2), ç ç , ÷ ÷ , (-3, -6) è3 3 ø è3 3ø

11

Conic Sections 1.

If the chord y = mx + 1 of the circle x2+y2=1 subtends an angle of measure 45° at the major segment of the circle then value of m is [2002] (a) 2 ± 2

2.

3.

4.

5.

7.

(b) x2 + y2 £ 25

æ1 1ö ç , ÷ è2 2ø

(b)

æ1 ö ç ,- 2 ÷ 2 è ø

(c)

æ3 1ö ç , ÷ è2 2ø

(d)

æ1 3ö ç , ÷ è2 2ø

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is [2002] (a) x2 + y2 = 9a2 (b) x2 + y2 = 16a2 (c) x2 + y2 = 4a2 (d) x2 + y2 = a2 Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002] (a) x = ± ( y + 2 a ) y = ± ( x + 2a)

(b) (c)

x = ± ( y + a)

(d)

y = ± ( x + a)

The lines 2 x - 3 y = 5 and 3x - 4 y = 7 are diameters of a circle having area as 154 sq.units.Then the equation of the circle is [2003] (a) x 2 + y 2 - 2 x + 2 y = 62 (b) x 2 + y 2 + 2 x - 2 y = 62

(d) 3 £ x2+ y2 £ 9 (c) x2 + y2 ³ 25 The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x2 + y2 = 9 is [2002] (a)

If the two circles ( x - 1) 2 + ( y - 3) 2 = r 2 and x 2 + y 2 - 8 x + 2 y + 8 = 0 intersect in two distinct point, then [2003] (a) r > 2 (b) 2 < r < 8 (c) r < 2 (d) r = 2.

(b) –2 ± 2

(c) –1 ± 2 (d) none of these The centres of a set of circles, each of radius 3, lie on the circle x2+y2=25. The locus of any point in the set is [2002] (a) 4 £ x2 + y2 £ 64

6.

(c) x 2 + y 2 + 2 x - 2 y = 47 (d) x 2 + y 2 - 2 x + 2 y = 47 . 8.

The normal at the point (bt12 , 2bt1 ) on a parabola meets the parabola again in the point (bt 2 2 , 2bt 2 ) , then

(a) t 2 = t1 +

2 t1

(c) t2 = -t1 + 9.

2 t1

The foci of the ellipse hyperbola

[2003] (b)

t 2 = -t1 -

(d)

t 2 = t1 -

2 t1

2 t1

x2 y 2 + = 1 and the 16 b2

x2 y 2 1 coincide. Then the = 144 81 25

value of b 2 is (a) 9

(b) 1

(c) 5

(d) 7

[2003]

EBD_7764 M-56 10. If a circle passes through the point (a, b) and

cuts the circle x 2 + y 2 = 4 orthogonally, then the locus of its centre is [2004]

11.

12.

13.

(a)

2ax - 2by - (a 2 + b 2 + 4) = 0

(b)

(a)

4 x2 + 3 y 2 = 1

2ax + 2by - (a 2 + b 2 + 4) = 0

(b)

3 x 2 + 4 y 2 = 12

(c)

2ax - 2by + (a 2 + b 2 + 4) = 0

(c)

4 x 2 + 3 y 2 = 12

(d)

2ax + 2by + (a 2 + b 2 + 4) = 0

(d)

3x 2 + 4 y 2 = 1

A variable circle passes through the fixed point A( p, q ) and touches x-axis . The locus of the other end of the diameter through A is [2004] (a) ( y - q)2 = 4 px

(b) ( x - q)2 = 4 py

(c) ( y - p)2 = 4qx

(d) ( x - p)2 = 4qy

If the lines 2 x + 3 y + 1 = 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference 10p, then the equation of the circle is [2004] (a)

x 2 + y 2 + 2 x - 2 y - 23 = 0

(b)

x 2 + y 2 - 2 x - 2 y - 23 = 0

(c)

x 2 + y 2 + 2 x + 2 y - 23 = 0

(d)

x 2 + y 2 - 2 x + 2 y - 23 = 0

Intercept on the line y = x by the circle x 2 + y 2 - 2 x = 0 is AB. Equation of the circle on AB as a diameter is [2004]

16. If the circles x 2 + y 2 + 2ax + cy + a = 0 and x 2 + y 2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for [2005] (a) exactly one value of a (b) no value of a (c) infinitely many values of a (d) exactly two values of a 17. A circle touches the x- axis and also touches the circle with centre at (0,3 ) and radius 2. The locus of the centre of the circle is [2005] (a) an ellipse (b) a circle (c) a hyperbola (d) a parabola 18. If a circle passes through the point (a, b) and

cuts the circle x 2 + y 2 = p 2 orthogonally, then the equation of the locus of its centre is [2005] (a)

x 2 + y 2 – 3ax – 4by + ( a 2 + b 2 - p 2 ) = 0

(b) 2ax + 2by – ( a 2 - b 2 + p 2 ) = 0

(a) x 2 + y 2 + x - y = 0

(c)

(b) x 2 + y 2 - x + y = 0

2 2 2 (d) 2ax + 2by – ( a + b + p ) = 0

(c) x 2 + y 2 + x + y = 0 (d) x 2 + y 2 - x - y = 0 14.

Mathematics 15. The eccentricity of an ellipse, with its centre at 1 the origin, is . If one of the directrices is x = 4, 2 then the equation of the ellipse is: [2004]

If a ¹ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y 2 = 4ax and x 2 = 4ay, then

(a) (b)

d 2 + (3b - 2c) 2 = 0

(c)

d 2 + (2b - 3c) 2 = 0

(d)

d 2 + (2b + 3c) 2 = 0

2

2

d + (3b + 2c) = 0

[2004]

x 2 + y 2 – 2ax – 3by + ( a2 - b2 - p 2 ) = 0

19. If the pair of lines ax 2 + 2 (a + b)xy + by 2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then [2005] (a)

3a 2 - 10 ab + 3b 2 = 0

(b)

3a 2 - 2ab + 3b 2 = 0

(c)

3a 2 + 10 ab + 3b 2 = 0

(d)

3a 2 + 2ab + 3b 2 = 0

M-57

Conic Sections

20.

Let P be the point ( 1, 0 ) and Q a point on the locus y = 8 x . The locus of mid point of PQ is [2005] (a)

(b) y 2 + 4x + 2 = 0

y 2 – 4x + 2 = 0 2

2

(c) x + 4y + 2 = 0

21.

(d) x – 4y + 2 = 0 The locus of a point P (a, b) moving under the condition that the line y = ax + b is a tangent to the hyperbola

22.

a2

-

y2 b2

= 1 is

[2005]

(a) an ellipse (b) a circle (c) a parabola (d) a hyperbola An ellipse has OB as semi minor axis, F and F ' its focii and the angle FBF ' is a right angle. Then the eccentricity of the ellipse is [2005] 1 1 (a) (b) 2 2 (c)

23.

x2

1 4

(d)

1 3

If the lines 3 x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 are two diameters of a circle of area 49p square units, the equation of the circle is [2006]

24.

25.

(a)

x 2 + y 2 + 2 x - 2 y - 47 = 0

(b)

x 2 + y 2 + 2 x - 2 y - 62 = 0

(c)

x 2 + y 2 - 2 x + 2 y - 62 = 0

(d)

x 2 + y 2 - 2 x + 2 y - 47 = 0

Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of 2p at its center is [2006] 3 (a) x 2 + y 2 = 3 (b) x 2 + y 2 = 1 2 9 27 2 2 (c) x + y = (d) x 2 + y 2 = 4 4 The locus of the vertices of the family of 3 2 2 parabolas y = a x + a x - 2a is

3

2

[2006]

3 105 (b) xy = 4 64 35 (c) xy = (d) xy = 64 16 105 In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is [2006] (a) 3 (b) 1 5 2 (c) 4 (d) 1 5 5

(a)

2

26.

xy =

27. Consider a family of circles which are passing through the point (– 1, 1) and are tangent to xaxis. If (h, k) are the coordinate of the centre of the circles, then the set of values of k is given by the interval [2007] 1 1 £k£ 2 2

(a)

-

(c)

0£ k £

1 2 x2

28. For the Hyperbola

(b)

k£

1 2

(d)

k³

1 2

y2

= 1 , which of cos 2 a sin 2 a the following remains constant when a varies = ? [2007] (a) abscissae of vertices (b) abscissae of foci (c) eccentricity (d) directrix. 29. The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is [2007] (a) (2, 4) (b) (–2, 0) (c) (–1, 1) (d) (0, 2) 30. The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is [2008] (a) (3, – 4) (b) (–3, 4) (c) (–3, –4) (d) (3, 4) 31. A focus of an ellipse is at the origin. The directrix -

is the line x = 4 and the eccentricity is the length of the semi-major axis is (a)

8 3

(b)

2 3

(c)

4 3

1 . Then 2 [2008]

(d)

5 3

EBD_7764 M-58 32. A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at [2008] (a) (0, 2) (b) (1, 0) (c) (0, 1) (d) (2, 0) 33. If P and Q are the points of intersection of the

circles x 2 + y2 + 3 x + 7 y + 2 p - 5 = 0 and x2 + y2 + 2x + 2y – p2=0 then there is a circle passing through P, Q and (1, 1) for: [2009] (a) all except one value of p (b) all except two values of p (c) exactly one value of p (d) all values of p 34.

The ellipse

x 2 + 4 y 2 = 4 is inscribed in a

rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is : [2009]

35.

36.

37.

38.

(a) x 2 + 12 y 2 = 16

(b) 4 x 2 + 48 y 2 = 48

(c) 4 x 2 + 64 y 2 = 48

(d) x 2 + 16 y 2 = 16

The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if [2010] (a) – 35 < m < 15 (b) 15 < m < 65 (c) 35 < m < 85 (d) – 85 < m < – 35 If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is [2010] (a) 2x + 1 = 0 (b) x=–1 (c) 2x – 1 = 0 (d) x= 1 2 2 2 2 The two circles x + y = ax and x + y = c2 (c > 0) touch each other if [2011] (a) | a | = c (b) a = 2c (c) | a | = 2c (d) 2 | a | = c The shortest distance between line y – x =1 and curve x = y2 is [2011] (a)

(c)

3 2 8 4 3

(b)

(d)

8 3 2 3 4

39.

Mathematics Equation of the ellipse whose axes are the axes of coordinates and which passes through the

2 is [2011] 5 (a) 5x2 + 3y2 – 48 = 0 (b) 3x2 + 5y2 – 15 = 0 (c) 5x2 + 3y2 – 32 = 0 (d) 3x2 + 5y2 – 32 = 0 The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is [2011 RS]

point (–3, 1) and has eccentricity

40.

(a)

x2 + y 2 - 2 x - 2 y + 1 = 0

(b) x2 + y2 – x – y = 0 (c) x2 + y2 + 2x + 2y – 7= 0 (d) x2 + y2 + x + y – 2 = 0 41. The equation of the hyperbola whose foci are (– 2, 0) and (2, 0) and eccentricity is 2 is given by : [2011RS] (a) x2 – 3y2 = 3 (b) 3x2 – y2 = 3 (c) – x2 + 3y2 = 3 (d) – 3x2 + y2 = 3 42. The length of the diameter of the circle which touches the x-axis at the point (1,0) and passes through the point (2,3) is: [2012] (a)

10 3

(b)

3 5

(c)

6 5

(d)

5 3

43. Statement-1 : An equation of a common tangent to the parabola y 2 = 16 3 x and the ellipse 2 x 2 + y 2 = 4 is y = 2x + 2 3

4 3 , (m ¹ 0) m is a common tangent to the parabola

Statement-2 : If the line y = mx +

y 2 = 16 3 x and the ellipse 2x2 + y2 = 4, then m

satisfies m4 + 2m2 = 24 [2012] (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false.

Conic Sections

44.

An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of the circle x2 + (y – 2)2 = 4 is semimajor axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : [2012] (a) 4x2 + y2 = 4 (b) x2 + 4y2 = 8

47.

(c) 4x2 + y2 = 8 (d) x2 + 4y2 = 16 The chord PQ of the parabola y2 = x, where one end P of the chord is at point (4, – 2), is perpendicular to the axis of the parabola. Then the slope of the normal at Q is [2012] 1 (a) – 4 (b) 4 1 (c) 4 (d) 4 The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point [2013] (a) (–5, 2) (b) (2, –5) (c) (5, –2) (d) (–2, 5) The equation of the circle passing through the

48.

x2 y 2 + = 1, and having 16 9 centre at (0, 3) is [2013] (a) x2 + y2 – 6y – 7 = 0 (b) x2 + y2 – 6y + 7 = 0 (c) x2 + y2 – 6y – 5 = 0 (d) x2 + y2 – 6y + 5 = 0 Given : A circle, 2x2 + 2y2 = 5 and a parabola, y2

45.

46.

foci of the ellipse

= 4 5 x. Statement-1 : An equation of a common tangent to these curves is y = x + 5 .

5 (m ¹ 0) is their m common tangent, then m satisfies m4 – 3m2 + 2 = 0. [2013] (a) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. Statement-2 : If the line, y = mx +

M-59 49. The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is [2014]

(a)

( x2 + y2 )

2

= 6 x2 + 2 y 2

(b)

( x2 + y2 )

2

= 6 x2 - 2 y 2

(c)

( x2 - y2 )

2

= 6x2 + 2 y2

(d)

( x2 - y 2 )

2

= 6 x2 - 2 y 2

50. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to [2014] (a)

1 2

(b) 3

1 4

3 2 2 51. The slope of the line touching both the parabolas (c)

(d)

y 2 = 4 x and x 2 = -32 y is

(a)

1 8

(b)

[2014] 2 3

1 3 (d) 2 2 52. Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then locus of P is : [2015] 2 2 (a) y = 2x (b) x = 2y (c) x2 = y (d) y2 = x 53. The number of common tangents to the circles x2 + y2 – 4x – 6x – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is : [2015] (a) 3 (b) 4 (c) 1 (d) 2 54. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera

(c)

recta to the ellipse

x 2 y2 + = 1, is : 9 5

[2015]

EBD_7764 M-60

(a)

27 2

(b) 27

(c)

27 4

(d) 18

Mathematics 58. If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is: [2016] (a) 5 (b) 10

55. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0, k Î R, is a : [2015] (a) circle of radius

2.

(b) circle of radius

3. (c) straight line parallel to x-axis (d) straight line parallel to y-axis 56.

57.

(a)

(c)

(b)

3 4 3

(d)

(d) 5 3 5 2 59. Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is: [2016] x + 2y - 24 = 0 4 (b) x2 + y2 – 4x + 9y + 18 = 0 (c) x2 + y2 – 4x + 8y + 12 = 0 (d) x2 + y2 – x + 4y – 12 = 0 60. A hyperbola passes through

(a)

The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis, lie on: [2016] (a) a hyperbola (b) a parabola (c) a circle (d) an ellipse which is not a circle The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : [2016]

2

(c)

3

P

(

x 2 + y2 -

the point

)

2, 3 and has foci at (± 2, 0). Then the

tangent to this hyperbola at P also passes through the point : [2017] (a) (c)

((2

) 3)

2, - 3

(b)

2,3

(d)

(3 2, 2 3 ) ( 3, 2 )

61. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is : [2017] (a)

4 3

(c)

( 2( 4

) 2 - 1) 2 +1

( 4(

(b) 2 (d)

) 2 - 1) 2 +1

Answer Key 1 2 (c) (a) 16 17 (b) (d) 31 32 (a) (b) 46 47 (c) (a) 61 (None)

3 (b) 18 (d) 33 (a) 48 (b)

4 (c) 19 (d) 34 (a) 49 (a)

5 (b) 20 (a) 35 (a) 50 (b)

6 (b) 21 (d) 36 (b) 51 (c)

7 (d) 22 (a) 37 (a) 52 (b)

8 (b) 23 (d) 38 (a) 53 (a)

9 (d) 24 (d) 39 (d) 54 (b)

10 (b) 25 (a) 40 (b) 55 (a)

11 (d) 26 (a) 41 (b) 56 (b)

12 (d) 27 (d) 42 (a) 57 (a)

13 (d) 28 (b) 43 (b) 58 (d)

14 (d) 29 (b) 44 (d) 59 (c)

15 (b) 30 (c) 45 (a) 60 (c)

M-61

Conic Sections

1.

(c) Equation of circle x2 + y2 = 1 = (1)2 Þ x2 + y2 = (y – mx)2 Þ x2 = m2x2 – 2 mxy; Þ x 2 (1 – m 2 ) + 2mxy = 0. Which represents the pair of lines between which the angle is 45o. tan 45 = ±

2 m2 - 0

=

±2 m

9 1 - =2 \f =± 2. 4 4 Hence, the centres of required circle are Þ f2 =

4. ;

1 - m2 1 - m2 Þ 1 – m2 = ± 2m Þ m2 ± 2m – 1 = 0

Þm=

ö ö æ1 æ1 ç , 2 ÷ or ç ,- 2 ÷ ø ø è2 è2 (c) Let ABC be an equilateral triangle, whose median is AD. A

O

-2 ± 4 + 4 2

B

-2 ± 2 2 = -1 ± 2 . 2 (a) For any point P (x, y) in the given circle,

Given AD = 3a.

=

2.

In D ABD, AB2 = AD2 + BD2 ; Þ x2 = 9a2 + (x2/4) where AB = BC = AC = x.

Y A C O

B P

3 2 x = 9a2 Þ x2 = 12a2. 4

X

In D OBD, OB2 = OD2 + BD2

Þ r2 = (3a – r)2 +

we should have OA £ OP £ OB Þ (5 - 3) £ x 2 + y 2 £ 5 + 3

3.

C

D

Þ 4 £ x 2 + y 2 £ 64 (b) Let the required circle be x2 + y2 + 2gx + 2fy + c = 0 Since it passes through (0, 0) and (1, 0)

5.

Þ r2 = 9a2 – 6ar + r2 + 3a2 ; Þ 6ar = 12a2 Þ r = 2a So equation of circle is x2 + y2 = 4a2 (b) Any tangent to the parabola y2 = 8ax is 2a ...(i) m If (i) is a tangent to the circle, x2 + y2 = 2a2

y = mx+

1 2 Points (0, 0) and (1, 0) lie inside the circle x2 + y2 = 9, so two circles touch internally Þ c1c2 = r1 – r2

Þ c = 0 and g = -

\ g 2 + f 2 = 3 - g 2 + f 2 Þ g2 + f 2 =

x2 4

then,

2a = ±

2a m m2 + 1

Þ m2(1 + m2) = 2 Þ ( m2 + 2)(m2 – 1) = 0 Þ m = ± 1. So from (i), y = ± (x + 2a). 3 6. 2

(b)

r1 - r2 < C1C2 for intersection Þ r -3< 5Þ r C1C2 , r + 3 > 5 Þ r > 2 ...(2) 7.

(d)

From (1) and (2), 2 < r < 8.

x 2 + y 2 + 2 gx + 2 fy + c = 0

pr 2 = 154 Þ r = 7 For centre on solving equation

It passes through (a, b) \ a 2 + b2 + 2 ga + 2 fb + c = 0

2 x - 3 y = 5& 3x - 4 y = 7 \ centre = (1, –1 )

\ 2( g ´ 0 + f ´ 0) = c - 4 Þ c = 4

Equation of circle, ( x - 1)2 + ( y + 1)2 = 7 2

\ from (2) a 2 + b2 + 2 ga + 2 fb + 4 = 0 \ Locus of centre (–g,–f) is

x 2 + y 2 - 2 x + 2 y = 47

a 2 + b2 - 2ax - 2by + 4 = 0

(b) Equation of the normal to a parabola y 2 = 4bx at point

(bt

2 1 , 2bt1

)

(bt

2 2 , 2bt 2

) then (

)

Þ 2 = – t1(t2 + t1) Þ t2 + t1 = – t 1 2 t1

144 ,b = 25

....(1)

\ p 2 + q 2 + 2 gp + 2 fq + c = 0

....(2)

....(3)

Let the other end of diameter through (p, q) be (h, k), then h+ p k+q = - g and =-f 2 2

Put in (3) 2

æ h + pö æ k + qö æ h + pö p2 + q 2 + 2 p ç ÷ + 2q ç ÷ +ç ÷ =0 è è 2 ø 2 ø è 2 ø

x2 y2 1 = 144 81 25 a=

x 2 + y 2 + 2 gx + 2 fy + c = 0

p 2 + q 2 + 2 gp + 2 fq + g 2 = 0 2

(d)

(d) Let the variable circle be

\ g 2 - c = 0 Þ c = g 2 . From (2)

= –t1(t2 + t1) (t2 – t1)

Þ t2 = – t1 –

11.

Circle (1) touches x-axis,

2bt2 = – t1 bt22 + 2 bt1 + bt13

2t2 – 2t1 = – t1 t22 – t12

or 2ax + 2by = a 2 + b 2 + 4

is

y = – t1 x + 2bt1 + bt13 As given, it also passes through

9.

......(2)

(1) cuts x 2 + y 2 = 4 orthogonally

we get x = 1, y = -1

8.

......(1)

81 81 15 5 , e = 1+ = = 25 144 12 4

\ \ \

Foci = ( ±3 , 0) foci of ellipse = foci of hyperbola for ellipse ae = 3 but a = 4,

\

3 e= 4

Then b 2 = a 2 (1 - e 2 ) Þ b 2 = 16æç1 - 9 ö÷ = 7 è 16 ø

Þ h2 + p 2 - 2hp - 4kq = 0

\ locus of (h, k) is x 2 + p2 - 2 xp - 4 yq = 0 Þ ( x - p )2 = 4qy

12. (d) Two diameters are along 2 x + 3 y + 1 = 0 and 3x - y - 4 = 0 solving we get centre (1, –1) circumference = 2pr = 10p \ r = 5.

M-63 \ The two equations should represent the same line

Conic Sections

Required circle is, ( x - 1)2 + ( y + 1)2 = 52 Þ x 2 + y 2 - 2 x + 2 y - 23 = 0

13.

Þ

(d) Solving y = x and the circle x 2 + y 2 - 2 x = 0, we get

x = 0, y = 0 and x = 1, y = 1

(d)

a2 + a + 1 = 0 No real value of a. (d) Equation of circle with centre (0, 3) and

\ Extremities of diameter of the required circle are (0, 0) and (1, 1). Hence, the equation of circle is

radius 2 is x 2 + ( y - 3)2 = 4

( x - 0)( x - 1) + ( y - 0)( y - 1) = 0

Q \

Þ x2 + y 2 - x - y = 0

14.

17.

a c - d a +1 Þ a + 1 = -a 2 = = 1 b -a

Solving equations of parabolas

Let locus of the variable circle is (a , b ) It touches x - axis. It¢s equation is

( x - a )2 + ( y + b)2 = b 2

y 2 = 4ax and x 2 = 4ay

we get (0, 0) and ( 4a, 4a) Substituting in the given equation of line

c1

r1

2bx + 3cy + 4d = 0,

r2

we get d = 0 and 2b +3c = 0

c2

(a , b )

Þ d 2 + (2b + 3c )2 = 0

15.

(b)

1 a e = . Directrix , x = = 4 2 e

1 \a = 4´ = 2 2 \b = 2 1-

1 = 3 4

Equation of ellipse is

16.

(b)

Circle touch externally Þ c1c2 = r1 + r2 \ a 2 + (b - 3) 2 = 2 + b

a 2 + (b - 3) 2 = b 2 + 4 + 4b

Þ a 2 = 10(b - 1 / 2)

x2 y 2 + = 1 Þ 3x 2 + 4 y 2 = 12 4 3

1 \ Locus is x 2 = 10 æç y - ö÷ è 2ø

s1 = x 2 + y 2 + 2ax + cy + a = 0

Which is parabola.

s2 = x 2 + y 2 - 3ax + dy - 1 = 0

Equation of common chord of circles s1 and s2 is given by s1 - s2 = 0

Þ 5ax + (c - d ) y + a + 1 = 0 Given that 5x + by – a = 0 passes through P and Q

18. (d) Let the centre be (a, b) 2 2 2 Q It cuts the circle x + y = p orthogonally

\ Using 2 g1 g2 + 2 f1 f 2 = c1 + c2 , we get

2(-a ) ´ 0 + 2(- b) ´ 0 = c1 - p2

EBD_7764 M-64

Mathematics

c1 = p 2

a=

Let equation of circle is x 2 + y 2 - 2ax - 2by + p 2 = 0

h +1 , 2

b=

k+0 2

2 b = k.

2 a -1 = h

It passes through

(2b) 2 = 8(2a - 1) Þ b 2 = 4a - 2

(a, b) Þ a 2 + b 2 - 2aa - 2bb + p 2 = 0

Þ y2 - 4x + 2 = 0 .

\ Locus of (a, b) is

\ 2ax + 2by - (a 2 + b 2 + p 2 ) = 0 .

21. (d) Tangent to the hyperbola y = mx ±

3q

(d)

As per question area of one sector = 3 area of another sector Þ angle at centre by one sector = 3 ´ angle at centre by another sector Let one angle be q then other = 3q Clearly q + 3q = 180 Þ q = 45o \ Angle between the diameters represented by combined equation ax 2 + 2 ( a + b ) xy + by 2 = 0 is 45o

\ Using tan q = we get tan 45o =

Þ1=

2 h2 - ab a+b 2

( a + b) 2 - ab

(

2

2

)

Þ a 2 + b 2 + 2 ab = 4a 2 + 4b 2 + 4ab 2

Þ 3a + 3b + 2 ab = 0

20.

b2

= 1 is

a 2 m2 - b2

Þ

m = a and a 2 m2 - b2 = b 2

\

a 2a 2 - b 2 = b 2

Locus is hyperbola.

a 2 x 2 - y 2 = b2 which

(a) P = (1, 0) Q = (h, k) Such that K 2 = 8h Let (a, b) be the midpoint of PQ

is

22. (a) Q ÐFBF ' = 90° Þ FB 2 + F ' B 2 = FF ' 2

(

\ a 2e2 + b2

) ( 2

a 2e 2 + b2

+

)

2

= (2ae) 2

b2 2 Þ 2(a 2 e 2 + b2 ) = 4a 2 e 2 Þ e = 2 a B (0, b)

F' (-ae, 0)

O

F (ae, 0)

Also e 2 = 1 - b 2 / a 2 = 1 - e2

Þ ( a + b) = 4 a + b + ab

2

y2

a+b

2 a 2 + b2 + ab a+b 2

a2

-

Given that y = a x + b is the tangent of hyperbola

q

19.

x2

Þ 2e 2 = 1, e =

1 2

.

23. (d) Point of intersection of 3 x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 is (1, - 1) which is the centre of the circle and radius = 7 \ Equation is ( x - 1)2 + ( y + 1) 2 = 49

Þ x 2 + y 2 - 2 x + 2 y - 47 = 0

M-65

Conic Sections

24.

(d)

Let M(h, k) be the mid point of chord AB 2p where ÐAOB = 3

3 3 = a 5 27. (d) Equation of circle whose centre is (h, k) i.e (x – h)2 + (y – k)2 = k2

\ e=

O (0, 0) 3 p/3 A

\ ÐAOM = Þ

(h, k)

M(h, k) B

p 3 p . Also OM = 3cos = 3 2 3

h2 + k 2 =

3 9 Þ h2 + k 2 = 2 4

2 2 \ Locus of (h, k) is x + y =

25.

(a) Given parabola is y =

9 4

a3 x 2 a 2 x + - 2a 3 2

9 ö 3a a3 æ 3 3 - 2a çè x + 2a x + ÷3 16a 2 ø 16

Þy=

Þy+

(-1,1)

3

35a a æ 3ö = çx+ ÷ è 16 3 4a ø

2

æ -3 -35a ö ÷ \ Vertex of parabola is çè , 4a 16 ø To find locus of this vertex, x=

-35a -3 and y = 4a 16

Þ a=

-3 16 y and a = 4x 35

-3 -16 y = Þ Þ 64xy = 105 4x 35

Þ xy = 26.

(a)

105 which is the required locus. 64

2ae = 6 Þ ae = 3 ; 2b = 8 Þ b = 4 2 2 2 b 2 = a 2 (1 - e 2 ) ; 16 = a - a e

Þ a 2 = 16 + 9 = 25 Þ a = 5

X'

X

(radius of circle = k because circle is tangent to x-axis) Equation of circle passing through (–1, +1) \ (–1 –h)2 + (1 – k)2 = k2 Þ 1 + h2 + 2h + 1 + k2 – 2k = k2 Þ h2 + 2h – 2k + 2 = 0 D³0 \ (2)2 – 4 × 1.(–2k + 2) ³ 0 Þ 4 – 4(–2k + 2) ³ 0 Þ 1 + 2k – 2 ³ 0 1 2 28. (b) Given, equation of hyperbola is Þ k³

x2

y2

=1 cos 2 a sin 2 a We know that the equation of hyperbola is

x2 a

2

-

-

y2 b

2

= 1 Here, a 2 = cos 2 a and

b 2 = sin 2 a

We know that, b 2 = a 2 (e2 - 1) Þ sin 2 a = cos 2 a (e 2 - 1) Þ sin 2 a + cos2 a = cos 2 a.e2 Þ e 2 = 1 + tan 2 a = sec 2 a Þ e = sec a

\ ae = cos a .

1 =1 cos a

EBD_7764 M-66

Mathematics

Co-ordinates of foci are (± ae, 0)

29.

Y

i.e. ( ± 1, 0) Hence, abscissae of foci remain constant when a varies. (b) Parabola y2 = 8x Y

æa ö çè - ae÷ø e

X´

y 2 = 8x

Y´ (2,0)

X' x+2=0

F

8 3 \ Semi major axis = 8/3 (b) Vertex of a parabola is the mid point of focus and the point

32.

Y

We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix. Point must be on the directrix of parabola Q equation of directrix x + 2 = 0 Þ x = –2 Hence the point is (–2, 0) 30.

31.

Q(a,b) C(–1, –2)

Centre (–1, –2) Let Q ( a, b) be the point diametrically opposite to the point P(1, 0), 0+b 1+ a = –2 then = –1 and 2 2 Þ a = –3, b = – 4 So, Q is (–3, –4) (a) Perpendicular distance of directrix from focus a = – ae = 4 e 1ö æ Þaç2 – ÷ = 4 è 2ø

X

OA

B

Y¢

x= 2

(c) The given circle is x2 + y2 + 2x + 4y –3 = 0

P(1,0)

x= a/e

Þa =

X

Y'

X

O S (ae, 0)

where directrix meets the axis of the parabola. Here focus is O(0, 0) and directrix meets the axis at B(2, 0) \ Vertex of the parabola is (1, 0) 33. (a) The given circles are S1 º x2 + y2 + 3x + 7y + 2p – 5 = 0....(1) S 2 º x2 + y2 + 2x + 2y – p2 = 0 ....(2) \ Equation of common chord PQ is S1 – S2 = 0

Þ L º x + 5 y + p2 + 2 p - 5 = 0 Þ Equation of circle passing through P and Q is S1 + l L = 0 Þ (x2 + y2 + 3x + 7y + 2p – 5) + l (x + 5y + p2 +2p – 5) = 0 As it passes through (1, 1), therefore (7 + 2p ) + l (2p + p2 + 1) = 0 Þ l =–

2p + 7 ( p + 1) 2

M-67

Conic Sections

which does not exist for p = – 1 2

34.

x y + =1 4 1 So A = (2, 0) and B= (0, 1) If PQRS is the rectangle in which it is inscribed, then P = (2, 1).

(a) The given ellipse is

x2

a a = cÞ a =c 2 2 (a) Shortest distance between two curve occurred along the common normal, so – 2t = – 1 Þ t = 1/2

So,

2

38.

y

y2

+ = 1 be the ellipse a 2 b2 circumscribing the rectangle PQRS.

Let

Q

B (0,1)

(t2, t)

P (2, 1)

x

O O R

A (2,0) (4,0) S

So shortest distance between them is Then it passed through P (2,1 ) 4 1 \ + = 1 ....(A) 2 a b2 Also, given that, it passes through (4, 0) 16 \ 2 + 0 = 1 Þ a 2 = 16 a Þ b2 = 4/3 [substituting a2 = 16 in eqn (A)]

35.

x2 y2 \ The required ellipse is + =1 16 4 / 3 or x2 + 12y2 =16 (a) Circle x 2 + y 2 - 4 x - 8 y - 5 = 0

Centre = (2, 4), Radius =

4 + 16 + 5 = 5

If circle is intersecting line 3 x - 4 y = m, at two distinct points. Þ length of perpendicular from centre to the line < radius

36. 37.

6 - 16 - m

< 5 Þ 10 + m < 25 5 Þ –25 < m + 10 < 25 Þ – 35 < m < 15 (b) The locus of perpendicular tangents is directrix i.e., x = - a; x = -1 (a) If the two circles touch each other, then they must touch each other internally.

Þ

x2 2

+

y2

=1 a b2 through (– 3,

39. (d) Let the ellipse be It passes 9 1 + 2 = 1 ..(i) 2 a b

3 2 8

1)

so

Also, b 2 = a 2 (1 - 2 / 5) ...(ii) 32 32 , b2 = Solving (i) and (ii) we get a2 = 3 5 So, the equation of the ellipse is Þ 5b2 = 3a 2

3x 2 + 5 y 2 = 32 40. (b) Circle whose diametric end points are (1,0) and (0,1) will be of smallest radius. Equation of this smallest circle is (x – 1) (x – 0) + (y – 0) (y – 1) = 0 Þ x2 + y2 – x – y = 0 41. (b) ae = 2 e=2 \ a =1

(

)

b 2 = a 2 e2 - 1

b = 1( 4 - 1) 2

b2 = 3

EBD_7764 M-68

Equation of hyperbola, Þ

x

2

a

2

-

y

2

b2

Mathematics On comparing (1) and (2), we get

=1

x2 y 2 =1 1 3

3 x2 - y 2 = 3 42. (a) Let centre of the circle be (1,h) [Q circle touches x-axis at (1,0)] Y

(1,h) (2,3) C B X

A(1,0)

44.

Let the circle passes through the point B (2,3) \ CA = CB (radius) 2 2 Þ CA = CB Þ (1 – 1)2 + (h – 0)2 = (1 – 2)2 + ( h – 3)2 Þ h2 = 1 + h2 + 9 – 6h 10 5 = Þ h= 6 3 43. (b) Given equation of ellipse is 2x2 + y2 = 4 2

2

2

2

y y 2x x + =1 Þ + =1 4 4 2 4 Equation of tangent to the ellipse Þ

x2 y 2 + = 1 is 2 4 ...(1) y = mx ± 2m2 + 4 (Q equation of tangent to the ellipse

4 3 = ± 2m 2 + 4 m Squaring on both the sides, we get 16 (3) = (2m2 + 4) m2 Þ 48 = m2 (2m2 + 4) Þ 2m4 + 4m2 – 48 = 0 Þ m4 + 2m2 –24 = 0 Þ (m2 + 6)(m2 – 4) = 0 Þ m2= 4 (Q m2 ¹ – 6) Þ m = ± 2 Þ Equation of common tangents are y = ± 2x ± 2 3 Thus, statement-1 is true. Statement-2 is obviously true. (d) Equation of circle is (x – 1)2 + y2 =1 Þ radius = 1 and diameter = 2 \ Length of semi-minor axis is 2. Equation of circle is x2 + (y – 2)2 = 4 = (2)2 Þ radius = 2 and diameter = 4 \ Length of semi major axis is 4 We know, equation of ellipse is given by x2

( Major axis ) Þ

x2 (4)

2

+

+ 2

y2

( Minor axis )2

y2

2 2 =1 Þ x +y =1 (2) 16 4 2

Þ x2 + 4y2 =16 45. (a) Point P is (4, –2) and PQ ^ x-axis So, Q = (4, 2)

Y

nt ange T Q

y2 + 2 =1 a b

x2

Normal

2

2

2

X

2

is y = mx + c where c = ± a m + b ) Now, Equation of tangent to the parabola

4 3 ...(2) y 2 = 16 3 x is y = mx + m (Q equation of tangent to the parabola a y2 = 4ax is y = mx + ) m

=1

P (4, – 2)

Equation of tangent at (4, 2) is

M-69

Conic Sections

yy1 = Þ 2y = Þy=

1 (x + x1) 2 1 (x + 2) Þ 4y = x + 2 2

Þ Focii = (± 7, 0)

x 1 + 4 2

1 4 \ Slope of normal = – 4 (c) Since circle touches x-axis at (3, 0) \ the equation of circle be (x – 3)2 + (y – 0)2 + ly = 0

So, slope of tangent =

46.

7 4 Now, radius of this circle = a2 = 16 Þ e=

48.

Now equation of circle is (x – 0)2 + (y – 3)2 = 16 x2 + y2 – 6y – 7 = 0 (b) Let common tangent be

5 m Since, perpendicular distance from centre of the circle to the common tangent is equal to radius of the circle, therefore

y = mx +

5 m 1+ m

A (3, 0) A (1, –2)

2

=

5 2

On squaring both the side, we get m2 (1 + m2) = 2 Þ m4 + m2 – 2 = 0 Þ (m2 + 2)(m2 – 1) = 0 Þ m= ±1

47.

As it passes through (1, –2) \ Put x = 1, y = –2 Þ (1 – 3)2 + (–2)2 + l(–2) = 0 Þ l=4 \ equation of circle is (x – 3)2 + y2 – 8 = 0 Now, from the options (5, –2) satisfies equation of circle. (a) From the given equation of ellipse, we have 9 a = 4, b = 3, e = 1 16

(

(Q m ¹ ± 2 )

)

y = ± x + 5 , both statements are

correct as m = ±1 satisfies the given equation of statement-2. 49. (a) Given equation of ellipse can be written as x2 y 2 + =1 6 2 Þ a2 = 6, b2 = 2 Now, equation of any variable tangent is

y = mx ± a 2 m2 + b2

...(i)

where m is slope of the tangent So, equation of perpendicular line drawn from centre to tangent is -x m Eliminating m, we get y=

...(ii)

( x4 + y 4 + 2 x2 y 2 ) = a 2 x2 + b2 y 2

EBD_7764 M-70

50.

Þ

( x 2 + y 2 )2 = a 2 x 2 + b2 y 2

Þ

( x2 + y 2 )2 = 6 x2 + 2 y 2

(b)

Mathematics ALTERNATIVE METHOD:

1 m Since this is also tangent to x2 = – 32y

Let tangent to y2 = 4x be y = mx +

C (0, y)

\

(1, 1)

1ö æ x 2 = -32 ç mx + ÷ è mø

Þ x2 + 32mx +

T

32 =0 m

Now, D = 0

æ 32 ö (32) 2 - 4 ç ÷ = 0 è mø

Equation of circle C º ( x - 1)2 + ( y - 1)2 = 1

Radius of T = | y | T touches C externally therefore, Distance between the centres = sum of their radii Þ

3 Þm =

52.

(b)

1 ...(i) m Equation of tangent of parabola (2) y = mx + 8m2 ...(ii) (i) and (ii) are identical y = mx +

Þ

1 1 1 = 8m 2 Þ m3 = Þ m = 2 m 8

1 2

Let P(h, k) divides Let any point Q on x2 = 8y is (4t, 2t2).

Þ (0 – 1)2 + (y –1)2 = (1 + |y|)2 Þ 1 + y2 + 1 – 2y = 1 + y2 + 2| y | 2 | y | = 1 – 2y

51.

Þ m=

OQ in the ratio 1 : 3

(0 - 1)2 + ( y - 1)2 = 1+ | y |

1 If y > 0 then 2y = 1 – 2y Þ y = 4 If y < 0 then –2y = 1 – 2y Þ 0 = 1 (not possible) 1 \ y= 4 (c) Given parabolas are y2 = 4x ...(i) 2 x = –32y ...(ii) Let m be slope of common tangent Equation of tangent of parabola (1)

4 32

1

P3

Q (4t, 2t2)

O

Then by section formula

t2 and h = t 2

Þ

k=

Þ

2k = h 2 Required locus of P is x2 = 2y

53.

(a)

x2 + y2 – 4x – 6y – 12 = 0

...(i)

Centre, C1 = (2, 3) Radius, r1 = 5 units x2 + y2 + 6x + 18y + 26 = 0

...(ii)

Centre, C2 = (–3, –9) Radius, r2 = 8 units C1C2 =

(2 + 3)2 + (3 + 9)2 = 13 units

r1 + r2 = 5 + 8 = 13

M-71

Conic Sections \ C1 C2 = r1 + r2

By symmetry area of quadrilateral = 4 × (Area DOAB) = 4 ´ 55.

(a)

C1

Intersection point of 2x – 3y + 4 = 0 and x – 2y + 3 = 0 is (1, 2)

A(2, 3)

C2 Therefore there are three common tangents. 54.

(b)

P

The end point of latus rectum of ellipse

x

2

a

2

+

(1, 2)

æ b = 1 in first quadrant is ç ae, ÷ aø è b

y

2ö

2

2

B(a, b)

and the tangent at this point intersects x-axis at

Since, P is the fixed point for given family of lines

æa ö çè , 0÷ø and y-axis at (0, a). e

So, PB = PA (a – 1)2 + (b – 2)2 = (2 – 1)2 + (3 – 2)2

x2 y 2 The given ellipse is + =1 9 5

(a – 1)2 + (b –2)2 = 1 + 1 = 2

Then a2 = 9, b2 = 5 Þ \

e = 1–

(x – 1)2 + (y – 2)2 = ( 2) 2

5 2 = 9 3

(x – a)2 + (y – b)2 = r2 Therefore, given locus is a circle with centre (1,

end point of latus rectum in first quadrant is L (2, 5/3)

2) and radius 56.

Equation of tangent at L is

Area of DOAB =

2.

(b)

2x y + =1 9 3

It meets x-axis at A (9/2, 0) and y-axis at B (0, 3) \

C(4, 4)

1 9 27 ´ ´3= 2 2 4

6 k P(h, k)

Y B

k

(0, 3) L(2,5/3)

C

27 = 27 sq. units. 4

O

S

D

A X (9/2, 0)

X

X' For the given circle, centre : (4, 4) radius = 6

6+ k =

2

(h - 4)

(h – 4)2 = 20k + 20

+ (k - 4) 2

EBD_7764 M-72

Mathematics Centre of new circle = P(2t2, 4t) = P(2, – 4)

\ locus of (h, k) is (x – 4)2 = 20(y + 1), which is a parabola. 57.

(a)

2b 2 =8 a

Radius = PC =

1 and 2b = (2ae) 2

Þ 4b2 = a 2 e2 Þ 4a 2 (e 2 - 1) = a 2 e 2

Þ 3e2 = 4 Þ e =

=2 2 \

(d)

(–3, 2)

Þ

5

x2 + y2 – 4x + 8y + 12 = 0

Equation of hyperbola is

S

5 2

Centre of S : O (–3, 2) centre of given circle A(2, –3) Þ OA = 5 2 Also AB = 5 (Q AB = r of the given circle) Þ Using pythagoras theorem in DOAB

r=5 3 (c)

a2

-

Minimum distance Þ perpendicular distance Eqn of normal at p111(2t2, 4t) y = –tx + 4t + 2t3 It passes through C(0, –6) Þ t3 + 2t + 3 = 0 Þ t = – 1

2 a

2

-

3 b2 -3

2 2

4 - b b2

=1

b2

=1

...(2)

=1

Þ b4 + b2 – 12 = 0 Þ (b2 – 3) (b2 + 4) = 0 Þ b2 = 3 b2 = – 4 For b2 = 3

(Not possible)

2 2 Þ a2 = 1 \ x - y = 1 1 3

Equation of tangent is

2x 3y =1 1 3

Clearly (2 2,3 3) satisfies it.

2

P (2t , 4t) C

y2

Hyperbola passes through ( 2, 3 )

A (2, –3)

\

59.

x2

foci is (±2, 0) Þ ae = 2 Þ a2e2 = 4 Since b2 = a2 (e2 – 1) b2 = a2 e2 – a2 \ a2 + b2 = 4 ...(1)

O

B

2

( )

60. (c) 58.

Equation of circle is : (x –2)2 + (y + 4) = 2 2

2 3

(2 – 0) 2 + (–4 + 6)2

61. (None) (Let the equation of circle be x2 + (y – k)2 = r2 It touches x – y = 0

M-73

Conic Sections

Þ \

0-k 2

Þ

4 – y + (y – k)2 =

k2 2

Þ

1y2 – y(2k + 1) +

k2 + 4= 0 2

It will give equal roots \ D = 0

0

Þ

\

It touches y = 4 – x2 as well Solving the two equations

Þ

æ k2 ö (2k + 1)2 = 4 ç + 4 ÷ 2 è ø

Þ

2k2 + 4k – 15 = 0

Þ

k=

-2 + 34 2

\

r=

k -2 + 34 = 2 2 2

=r

k= r 2 Equation of circle becomes x2 + (y – k)2 =

k2 2

...(i)

Which is not matching with any of the option given here.1

EBD_7764 M-74

Mathematics

12

Limits and Derivatives 1.

[2002]

1 - cos 2 x

lim x ®0

is

2x

(a) 1 (c) zero 2.

3.

[2002] (b) –1 (d) does not exist

æ x2 + 5 x + 3 ö Lim ç ÷ x ®¥ çè x 2 + x + 2 ÷ø

(b) e2

(c) e3

(d) 1

Let f (x) = 4 and

f ¢ (x) = 4.

Then

(a) 2 (c) – 4

5.

(a)

1 p +1

(c)

1 1 p p -1

n p +1

8.

lim x®

[2002]

p 2

(b) 0 (d)

[2002]

(b)

1 1- p

(d)

1 p+2

9.

1 8 1 (d) 32

f ( x) - 1 x -1

2x

= e2 , then the values of

a and b, are

[2004]

(a) a = 1 and b = 2 (c)

log x n - [ x] lim , n Î N , ([x] denotes greatest [ x] x ®0

If f (1) = 1, f ¢(1) = 2, then lim x ®1

lim æ a b ö If x®¥ çç1+ + ÷÷ è x x2 ø

[2003]

(b)

(c) 0 is

2 3

é æ xö ù ê1 - tan çè 2 ÷ø ú [1 - sin x] ë û is é ù x æ ö 3 + p 1 tan [ 2 x ] çè ÷ø ú ê 2 û ë

(a) ¥

integer less than or equal to x) [2002] (a) has value -1 (b) has value 0 (c) has value 1 (d) does not exist 6.

log(3 + x ) - log(3 - x) = k , the value of x x®0 k is [2003]

If lim

(a) -

(b) –2 (d) 3

1 p + 2 p + 3 p + ..... + n p

(b) 4 (d) 1/2

2 3 1 (c) 3

[2002]

xf (2) - 2 f ( x) lim is given by x®2 x-2

4.

7.

x

(a) e4

lim n ®¥

(a) 2 (c) 1

is

a Î R, b = 2

(b)

a = 1, b Î R

(d)

a Î R, b Î R

10. Let a and b be the distinct roots of 2 ax 2 + bx + c = 0 , then lim 1 - cos(ax + bx + c ) 2

is equal to (a) (c)

a2 (a - b ) 2 2 -a 2 (a - b ) 2 2

( x - a)

x®a

[2005]

(b) 0 (d)

1 (a - b) 2 2

M-75

Limits and Derivatives

11.

(a) 0 (c) 2

Let f : R ® R be a positive increasing function with lim

x ®¥

f (3x ) f (2 x) = 1 then lim = f ( x) x ®¥ f ( x)

lim

14.

x®0

[2010] (a)

2 3

(b)

(c) 3 12.

(a)

æ 1 - cos{2( x - 2)} ö lim ç ÷÷ x ®2 ç x-2 è ø

(c) equals

[2011] (b) equals –

2

1

15.

2

16.

Let f : R ® [0, ¥) be such that lim f(x) exists x® 5 and lim x® 5

lim

( f ( x ) )2 - 9 = 0

1 4

1 2 (d) 2

(b)

(

sin p cos 2 x x2

) is equal to:

[2014]

(a)

-p

(b)

(c)

p 2

(d) 1

lim

p x® 2

[2011RS]

x -5

-

x®0

(d) does not exist

2

(1 - cos 2 x )(3 + cos x ) is equal to [2013] x tan 4 x

(c) 1

(d) 1

(a) equals

13.

3 2

(b) 1 (d) 3

f(x) equals : Then lim x® 5

cot x - cos x

( p - 2x )3

p

equals :

[2017]

(a)

1 4

(b)

1 24

(c)

1 16

(d)

1 8

Answer Key 1 (d) 16 (c)

1.

2 (a)

3 (c)

4 (a)

5 (d)

6 (a)

7 (d)

8 (d)

9 (b)

(d) lim

lim x®0

1 - cos 2 x 2x

2 sin 2 x 2x

Þ lim

1 - (1 - 2 sin 2 x) 2x

| sin x | Þ lim x®0 x

The limit of above does not exist as LHS = –1 ¹ RHL = 1

2. ;

(a)

10 (a)

11 (d)

12 (d)

æ x 2 + 5 x + 3ö lim ç 2 ÷ x ®¥ è x + x + 2 ø

13 (d)

14 (d)

15 (b)

x

æ 4x +1 ö = lim ç1 + 2 ÷ x ®¥ è x + x + 2ø

x

( 4 x +1) x

é x2 + x + 2 ù x2 + x + 2 êæ 4 x + 1 ö 4 x+1 ú ú = lim êç1 + 2 ÷ x ®¥ ê è x + x + 2ø ú ëê ûú

EBD_7764 M-76

Mathematics 4x2 + x

lim

=

2 e x®¥ x + x+ 2

4+ lim

1 x

x ®¥ 1+ 1 + 2 x 2

3.

æ yö tan ç - ÷ .(1 - cos y ) è 2ø = lim y®0 ( -2 y )3 y y - tan 2sin 2 2 2 = lim y®0 y3 (-8). .8 8

1 é lù êQ lim (1 + lx ) x = e ú ë x®¥ û

x =e =e (c) Apply L H Rule

4

xf (2) - 2 f ( x ) x-2 x®2

We have, lim

æ 0ö çè ÷ø 0

y 2 tan é 1 sin y / 2 ù 1 2 = lim .ê = ú y 32 y ®0 32 æ ö ë y / 2 û ç ÷ è2ø

= lim f (2) - 2 f ¢( x ) = f (2) - 2 f ¢ (2) x® 2

1

= 4 – 2 ´ 4 = –4. 4.

(a) We have lim n® ¥

9. 1p + 2 p + .... + n p n p+1

;

1

1 é x p +1 ù 1 p x dx = = ê ú = å ò p p + 1 +1 n ®¥ r =1 n p × n ëê ûú0 0 n

lim

5.

rp

(d) Since lim [ x ] does not exist, hence the x ®0

required limit does not exist. 6.

(a)

æ0ö ç ÷ form using L’’ lim è0ø x ®1 x -1 1 f ¢ ( x) 2 f ( x) Hospital’s rule = lim x ®1 1/ 2 x f ( x) - 1

= 7.

(d)

2 = =2. f (1) 1

log(3 + x) - log(3 - x) lim =k x ®0 x

lim x®

p 2

æ a bö \ lim ç1 + + ÷ x x2 ø x ®¥ è

æ p xö tan ç - ÷ .(1 - sin x ) è 4 2ø

Þ

é bù lim 2 ê a + ú xû

e x®¥ ë

= e2 æa b ö 2xç + ÷ è x x2 ø

2a 2 = e2 Þ e = e

Þ a = 1 and b Î R

Þe

a bö + x x 2 ÷ø

2x

= e2

ö lim æ a b x ®¥ ç 1+ + 2 -1÷ 2 x è x ø x

= e2

2b ö æ Þlim ÷=2 x®¥ ç 2a + è xø

Þ 2a + 0 = 2, b Î R Þ a = 1, b Î R 10. (a) Given limit = lim

1 - cos a ( x - a )( x - b )

x ®a

( p - 2 x )3

p Let x = + y; y ® 0 2

2x

æ 1 öù é êæ a b ö ç a + b ÷ ú Þ lim ê ç1 + + ÷ çè x 2 ÷ø ú x x x2 ø x®¥ è ê ú ë û

è

-1 1 + -x =k \2 = k 3 3 x Þ lim 1 3 x ®0

(d)

x ®¥

lim æ x ®¥ ç 1 +

f ¢ (1)

(by L 'Hospital rule)

8.

(b) We know that lim (1 + x ) x = e

= lim

x® a

( x - a )2 æ ( x - a )( x - b) ö 2sin 2 ç a ÷ø è 2 ( x - a )2

= e2

M-77

Limits and Derivatives

=

2

lim

x ® a ( x - a)2

´

x (1 - cos 2 x ) (3 + cos x) · 2 x®0 1 tan 4 x x lim

æ ( x - a ) ( x - b) ö sin 2 ç a ÷ø è 2

2sin 2 x 3 + cos x x · · 2 x®0 1 tan 4x x

a 2 ( x - a ) 2 ( x - b) 2 4

= lim

a 2 ( x - a )2 ( x - b)2 ´ 4

= 11.

= 2 lim

x®0

a 2 (a - b)2 . 2

15. (b) Consider lim

(d)

= lim sin

f (2 x ) =1 Þ xlim ®¥ f ( x)

sin q]

x ®0

= lim sin x ®0

1 - cos{2( x - 2)} x-2

x ®2

x®2

2 sin( x - 2) = -1 ( x - 2)

L.H.L = - lim

x®2

R.H.L = lim

(at x = 2)

x®2

2 sin( x - 2) =1 ( x - 2)

Thus L.H.L ¹ R.H.L (at x = 2)

Hence, lim

x® 2

1 - cos{2( x - 2)} does not x-2

exist. 13. (d)

lim

x® 5

x -5

=0

14.

lim

x2

[Q sin (p – q) =

(p sin 2 x) p sin 2 x ´ x2 p sin 2 x

cot x(1 - sin x)

p x® 2

pö æ -8 ç x - ÷ 2ø è

= limp

x®5

(d) Multiply and divide by x in the given expression, we get

3

cot x(1 - sin x) 3

æp ö 8ç - x ÷ 2 è ø p p Put - x = t Þ as x ® Þ t ® 0 2 2 æp öæ æ p öö cot ç - t ÷ ç 1 - sin ç - t ÷ ÷ è2 øè è 2 øø = lim 3 t®0 8t 2

= lim

lim [(f (x) )2 – 9] = 0 Þ lim f (x) = 3 x® 5

16. (c)

x®

(at x = 2)

( f ( x)) 2 - 9

(p - p sin 2 x)

2

x-2

(at x = 2)

x2

æ sin x ö = lim 1´ p ç ÷ =p x ®0 è x ø

2 sin( x - 2)

= lim

sin( p cos 2 x)

x2

x ®0

By Sandwich Theorem.

lim

x®0

sin éëp(1 - sin 2 x) ùû

= lim

f (2 x ) f (3 x ) 1 £ lim £ lim Þ xlim ®¥ x ®¥ f ( x) x ®¥ f ( x)

x x ®0 tan 4 x

· lim 3 + cos x · lim

x®0

f (2 x ) f (3 x) Þ 0 < 1 < f ( x) < f ( x)

12.

x

2

1 4x 1 lim = 2.4. = 2 4 x®0 tan 4 x 4

= 2.4

(d) f(x) is a positive increasing function. \ 0 < f ( x ) < f (2 x ) < f (3 x )

sin 2 x

tan t(1 - cos t)

8t 3 tan t 1 - cos t . = lim t ®0 8t t2 1 1 1 = .1. = 8 2 16 t®0

EBD_7764 M-78

Mathematics

Mathematical Reasoning DIRECTIONS: Given below question contains two statements: Statement-1(Assertion) and Statement-2(Reason). This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. 1. Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “ x is a rational number iff y is a transcendental number”. [2008] Statement-1 : r is equivalent to either q or p Statement-2 : r is equivalent to ~(p«~q). (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false 2. The statement p ® (q®p) is equivalent to [2008] (a) p ® (p® q) (b) p ® (p Ú q) (c) p ® (p Ù q) (d) p ® (p «q) DIRECTIONS: Given below question contains two statements: Statement-1(Assertion) and Statement 2(Reason). This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. 3.

Statement-1 : ~ ( p «~ q) is equivalent to p «q. Statement-2 : ~ ( p «~ q) is a tantology [2009] (a) Statement-1 is true, Statement-2 is true;

4.

5.

6.

13

Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -1 Let S be a non-empty subset of R. Consider the following statement : P : There is a rational number x Î S such that x > 0. Which of the following statements is the negation of the statement P ? [2010] (a) There is no rational number x Î S such than x < 0. (b) Every rational number x Î S satisfies x < 0. (c) x Î S and x < 0 Þ x is not rational. (d) There is a rational number x Î S such that x < 0. Consider the following statements [2011] P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as (a)

~ (Q « ( P Ù ~ R))

(b)

~ Q «~ P Ù R

(c)

~ ( P Ù ~ R) « Q

(d) ~ P Ù (Q «~ R ) The only statement among the following that is a tautology is [2011RS] (a)

A Ù (A Ú B)

(b)

A Ú (A Ù B)

M-79

Mathematical Reasoning

(c) 7.

8.

[A Ù (A ® B)] ® B

9.

(d) B ® [A Ù (A ® B)] The negation of the statement "If I become a teacher, then I will open a school", is : [2012] (a) I will become a teacher and I will not open a school. (b) Either I will not become a teacher or I will not open a school. (c) Neither I will become a teacher nor I will open a school. (d) I will not become a teacher or I will open a school. Consider Statement-1 : (p ^ ~ q) ^ (~ p ^ q) is a fallacy. Statement-2 : (p ® q) « (~ q ® ~ p) is a tautology. [2013] (a) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true.

The statement : ( p « : q ) is:

[2014]

(a) a tautology (b) a fallacy (c) equavalent to p « q (d) equivalent to : p « q 10. The negation of ~ s Ú (~ r Ù s) is equivalent to : [2015] (a)

s Ú (r Ú ~ s)

(c) s Ù ~ r 11. The Boolean Expression

(b)

sÙ r

(d)

s Ù (r Ù ~ s)

(p Ù : q) Ú qÚ (: p Ù q) is equivalent to: [2016] p Ú q p Ú : q (a) (b) (c) : p Ù q (d) p Ù q 12. The following statement [2017] (p ® q) ® [(~p ® q) ® q] is : (a) a fallacy (b) a tautology (c) equivalent to ~ p ® q (d) equivalent to p ® ~q

Answer Key 1 2 (None) (b)

1.

3 (b)

4 (b)

5 (a)

6 (c)

7 (a)

(None) p : x is an irrational number q : y is a transcendental number r : x is a rational number iff y is a transcendental number. clearly r : : p « q Let us use truth table to check th e equivalence of ‘r’ and ‘q or p’; ‘r’ and : ( p «: q) p T T F F

q T F T F

~p F F T T

~q F T F T

1 2 3 ~p «q q or p p«~q ~(p«~q) F T F T T T T F T T T F F F F T

8 (b)

9 (c)

2.

10 (b)

11 (a)

12 (b)

From columns (1), (2) and (3), we observe, none of the these statements are equivalent to each other. \ Statement 1as well as statement 2 both are false. \ None of the options is correct. (b) Let us make the truth table for the given statements, as follows : p T T F F

q pÚq T T F T T T F F

q®p T T F T

p ®(q®p) T T T T

p ®(pÚq) T T T T

From table we observe p ® (q®p) is equivalent to p®(pÚq)

EBD_7764 M-80

3.

(b) The truth table for the logical statements, involved in statement 1, is as follows : p

q

: q p « : q : ( p « : q) p « q

T T

F

F

T

T

T F F T

T F

T T

F F

F F

F

T

F

T

T

F

4.

We observe the columns for ~ (p « ~q) and p « q are identical, therefore ~(p « ~q) is equivalent to p « q 6.

7.

8.

9.

5.

(c)

Mathematics But ~ (p « ~q) is not a tautology as all entries in its column are not T. \ Statement-1 is true but statement-2 is false. (b) P : there is a rational number x Î S such that x > 0 ~ P : Every rational number x Î S satisfies x£0 (a) Suman is brilliant and dishonest if and only if Suman is rich is expressed as Q « ( P Ù ~ R) Negation of it will be ~ (Q « ( PÙ ~ R))

A

B A Ú B A Ù B A Ù (A Ú B) A Ú (A Ù B) A ® B A Ù (A ® B) [A Ù (A ® B) ®B]

T

F

T

F

T

T

F

F

T

T

F

T

T

F

F

F

T

F

T

F

T

T

T

T

T

T

T

T

T

T

F

F

F

F

F

F

T

F

T

T

\ It is tautology. (a) Let p : I become a teacher. q : I will open a school Negation of p ® q is ~ (p ® q) = p ^ ~q i.e. I will become a teacher and I will not open a school. (b) Statement-2 : (p ® q) « (~q ® ~p) º (p ® q) « (p ® q) which is always true. So statement 2 is true Statement-1: (p ^ ~q) ^ (~p ^ q) = p ^ ~q ^ ~p ^ q = p ^ ~p ^ ~q ^ q = f^f=f So statement-1 is true (c)

12. (b)

p F F

q F T

~q T F

T T

F T

T F

10. (b) :[:sÚ(:r Ù s)] = sÙ:(:r Ù s) = sÙ(r Ú:s) = (s Ù r) Ú (s Ù: s) = (s Ù r) Ú 0 =sÙr 11. (a)

(pÙ : q) Ú q Ú (: p Ù q) 40.

(1) (pÙ : q) Ú q Ú (: p Ù q)

Þ {(p Ú q) Ù (: q Ú q)} Ú (: p Ù q) Þ {(p Ú q) Ù T} Ú (: p Ù q) Þ (p Ú q) Ú (: p Ù q)

Þ {(p Ú q) Ú : p} Ù (p Ú q Ú q) Þ TÙ (p Ú q)

F T

Þ pÚq

We have p q

\

Clearly equivalent to p « q

p « ~ q ~ ( p « ~ q) F T T F T F

~ p p ® q ~ p ® q (~ p ® q) ® q (p ® q) ® ((~p ® q)® q)

T F F

F

T

F

T

T T F F F T

T T

T F

T T

T T

F T T

T

T

T

T

It is tautology.

[B ® [A Ù (A ®B)]

Statistics 1.

2.

3.

4.

5.

In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? [2002] (a) 73 (b) 65 (c) 68 (d) 74 The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set [2003] (a) remains the same as that of the original set (b) is increased by 2 (c) is decreased by 2 (d) is two times the original median. In an experiment with 15 observations on x, the following results were available: [2003] Sx 2 = 2830, Sx = 170 One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is [2003] (a) 8.33 (b) 78.00 (c) 188.66 (d) 177.33 Consider the following statements : (A) Mode can be computed from histogram (B) Median is not independent of change of scale (C) Variance is independent of change of origin and scale. Which of these is / are correct ? [2004] (a) (A), (B) and (C) (b) only (B) (c) only (A) and (B) (d) only (A) In a series of 2 n observations, half of them equal a and remaining half equal –a. If the standard deviation of the observations is 2, then |a| equals. [2004]

(a) (c) 2

2 n

(b) (d)

2

1 n

14

6.

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [2005] (a) 22.0 (b) 20.5 (c) 25.5 (d) 24.0

7.

Let x1 , x 2 , .............. xn be n observations such

8.

2 that å xi = 400 and å xi = 80. Then the possible value of n among the following is [2005] (a) 15 (b) 18 (c) 9 (d) 12 Suppose a population A has 100 observations 101, 102, ............., 200 and another population B has 100 obsevrations 151, 152, ................ 250. If VA and VB represent the variances of the two

populations, respectively then (a) 1

10.

9 4

4 2 (d) 9 3 The average marks of boys in class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is [2007] (a) 80 (b) 60 (c) 40 (d) 20. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b? [2008] (a) a = 0, b = 7 (b) a = 5, b = 2 (c) a = 1, b = 6 (d) a = 3, b = 4

(c)

9.

(b)

VA is [2006] VB

EBD_7764 M-82

11.

DIRECTIONS:This question contains two statements: statement-1 (Assertion) and statement-2 (Reason). This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. Statement-1 : The variance of first n even natural

16.

n2 –1 . 4 Statement-2 : The sum of first n natural numbers

numbers is

is

n(n + 1) and the sum of squares of first n 2

natural numbers is n (n + 1)(2n + 1) . [2009] 6 (a) Statement-1 is true, Statement-2 is true

12.

13.

Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true. Statement-2 is a correct explanation for Statement-1. If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to : [2009] (a) 20.0 (b) 10.1 (c) 20.2 (d) 10.0 For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is [2010] (a)

11 2

17.

18.

19.

(b) 6

13 5 (d) 2 2 If the mean deviation about the median of the numbers a, 2a,.......,50a is 50, then | a | equals [2011] (a) 3 (b) 4 (c) 5 (d) 2 A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standarion deviation of 2 gm. Later, it was found that the measuring scale was misaligned and

(c)

14.

15.

20.

Mathematics always under reported every fish weight by 2 gm. The correct mean and standard deviation (in gm) of fishes are respectively : [2011RS] (a) 32, 2 (b) 32, 4 (c) 28,2 (d) 28, 4 Let x1 , x2,...., xn be n observations, and let x be their arithmetic mean and s2 be the variance. Statement-1 : Variance of 2x1, 2x2, ..., 2xn is 4s2. Statement-2 : Arithmetic mean 2x1, 2x2, ..., 2xn is 4 x . [2012] (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ? [2013] (a) mean (b) median (c) mode (d) variance The variance of first 50 even natural numbers is [2014] 437 (a) 437 (b) 4 833 (c) (d) 833 4 The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is: [2015] (a) 15.8 (b) 14.0 (c) 16.8 (d) 16.0 If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true? [2016] 2 (a) 3a – 34a + 91 = 0 (b) 3a2 – 23a + 44 = 0 (c) 3a2 – 26a + 55 = 0 (d) 3a2 – 32a + 84 = 0

M-83

Statistics

Answer Key 1 (b) 16 (d)

1.

2 (a) 17 (d)

3 (b) 18 (d)

4 (c) 19 (b)

5 (c) 20 (d)

6 (d)

7 (b)

8 (a)

(b) Total student = 100; for 70 students total marks = 75 ´ 70 = 5250 Þ Total marks of girls = 7200 – 5250 = 1950 Average of girls =

9 (a)

6. 7.

1950 = 65 30

2.

3.

(b)

Sx = 170, Sx 2 = 2830 increase in

Þ

8.

Increase in Sx 2 = 900 - 400 = 500 then Sx' = 2830 + 500 = 3330

5.

9.

1 æ1 ö ´ 3330 - ç ´ 180 ÷ = 222 - 144 = 78. 15 è 15 ø

(c) Only first (A) and second (B) statements are correct. (c) Clearly mean A = 0 Standard deviation s = 2=

=

a 2 .2 n =| a | 2n

Hence | a | = 2

å x12 ³ æç å x1 ö÷ n

è

s 2x =

14 (b)

15 (a)

n ø

2

Þ

400 æ 80 ö ³ç ÷ è nø n

2

å di2

(Here deviations are taken n from the mean). Since A and B both have 100 consecutive integers, therefore both have same standard deviation and hence

å di2

is same in both the cases)

(a) Let the number of boys be x and that of girls be y. Þ 52x + 42y = 50(x + y) Þ 52x – 50x = 50y – 42y Þ 2x = 8y Þ

å ( x - A)2

( a - 0)2 + ( a - 0) 2 + ...(0 - a ) 2 + ... 2n

(a)

(As

2

2

4.

13 (a)

the variance. \ V A = 1 VB

2

=

12 (b)

Þ n ³ 16 . So only possible value for n = 18 Þ Mode = 3 × 22 – 2 × 21 = 66 – 42 = 24.

Sx = 10 , then Sx' = 170 + 10 = 180

1 2 æ1 ö Variance = Sx' -ç Sx' ÷ n èn ø

11 (c)

(d) Mode + 2Mean = 3 Median (b) We know that for positive real numbers x1, x2, ...., xn, A.M. of kth powers of x'i s ³ kth the power of A.M. of x'i s

th

æ 9 +1ö th (a) n = 9 then median term = ç ÷ =5 è 2 ø term. Last four observations are increased by 2. The median is 5th observation which is remaining unchanged. \ there will be no change in median.

10 (d)

x 4 x 4 = = and x+ y 5 y 1

Required % of boys =

2n

10.

x ´ 100 = x+ y

4 ´ 100 = 80 % 5 (d) Mean of a, b, 8, 5, 10 is 6 a + b + 8 + 5 + 10 = 10 5 Þ a+b=7

Þ

...(i)

EBD_7764 M-84

Mathematics

Variance of a, b, 8, 5, 10 is 6.80 (a - 6)2 + (b – 6) 2 + (8 – 6) 2 + (5 – 6) 2 + (10 – 6) 2 Þ 5

= 6.80 a 2 –12a + 36 + (1 – a ) 2 + 21 = 34 [using eq. (i)] Þ 2a2 –14a + 24 = 0 Þ a2 – 7a + 12 = 0 Þ a = 3 or 4 Þ b = 4 or 3 \ The possible values of a and b are a = 3 and b= 4 or, a = 4 and b = 3 (c) For the numbers 2, 4, 6, 8, ......., 2n 2[ n ( n + 1)] x = = (n + 1) 2n S ( x – x )2 S x2 – ( x )2 = And Var = 2n n

Þ 13. (a)

4S n 2 == – (n + 1)2 n

4n (n + 1) (2n + 1) = – (n + 1)2 6n 2(2n + 1) ( n + 1) – ( n + 1)2 = 3 é 4n + 2 – 3n – 3 ù = (n + 1) ê úû 3 ë

=

( n + 1)( n –1) = n2 - 1

3 3 \ Statement-1 is false. Clearly, statement 2 is true.

12.

101 + d(1 + 2 + 3 + ......+100) 101

(b)

Mean =

Q

d × 100 × 101 =1 + 50 d 101 × 2 Mean deviation from the mean = 255 =1+

Þ

1 [| 1 - (1 + 50d ) | + | (1 + d ) - (1 + 50 d ) | 101

+ | (1 + 2d )-(1 + 50d ) | +....+ | (1 + 100 d ) - (1 + 50 d ) |]= 255 Þ

2d [1 + 2 + 3 + ... + 50] = 101´ 255

Þ

2d ´

50 ´ 51 = 101´ 255 2

101´ 255 = 10.1 50 ´ 51

s 2x = 4, s 2y = 5, x = 2, y = 4

1 1 xi2 - (2)2 = 4; å yi2 - (4)2 = 5 å 5 5

Þ

11.

d=

å xi2 = 40; å yi2 = 105 2 2 Þ å ( xi + yi ) = 145 Þ å ( xi + yi ) = 5(2) + 5(4) = 30 Variance of combined data

(

)

1 æ1 ö = å xi2 + yi2 - çè 10 å ( xi + yi ) ÷ø 10

2

145 11 -9 = 10 2 (b) Median is the mean of 25th and 26th observation =

14.

\ M =

25a + 26a = 25.5a 2

M .D (M ) =

å xi - M N

1 [2 ´ a ´ (0.5 + 1.5 + 2.5 + ....24.5)] 50 25 Þ 2500 = 2 a ´ (25) 2 Þ 50 =

Þ a =4

15. (a) Correct mean = observed mean + 2 = 30 + 2 = 32 Correct S. D. = observed S.D. = 2 16. (d) A.M. of 2x1, 2x2, ..., 2xn is 2 x1 + 2x2 + ... + 2xn n

æ x + x2 + .... + xn ö =2ç 1 ÷ = 2x n è ø æ sum of observations ö çQ Mean = ÷ Number of observations ø è So statement-2 is false. variance (2xi) = 22 variance (xi) = 4s2 where i = 1, 2,......n So statement-1 is true.

M-85

Statistics

17.

(d) If initially all marks were xi then s12 =

å ( xi - x )2 i

N Now each is increased by 10 å [ ( xi +10) -( x +10) ]2

2

= i = s12 N N Hence, variance will not change even after the grace marks were given. (d) First 50 even natural numbers are 2, 4 , 6 ....., 100

s12 =

18.

i

å ( xi - x )

å

Variance = Þ

s2 =

xi2

N 2

2

2 + 4 + ... + 100 50

=

2

2

s=

- ( x) 2

æ 2 + 4 + ... + 100 ö -ç ÷ø è 50 2

æ 50 ´ 51 ´ 101ö = 4ç - (51)2 = 3434 – 2601 è 50 ´ 6 ÷ø Þ s2 = 833 19. (b) Sum of 16 observations = 16 × 16 = 256 Sum of resultant 18 observations = 256 – 16 + (3 + 4+5) = 252 252 = 14 Mean of observations = 18 2 + 3 + a + 11 a = +4 20. (d) x = 4 4

2

2

Þ3.5=

Þ

2

4(1 + 2 + 3 + .... + 50 ) - (51)2 50

å

x i2 - x n

( )

2

2 4 + 9 + a 2 +121 æ a ö - çç + 4 ÷÷ 4 è4 ø

49 4(134 + a 2 ) - (a 2 + 256 + 32a) = 4 16

Þ 3a 2 - 32a + 84 = 0

EBD_7764 M-86

Mathematics

15

Probability 1.

2.

A and B are events such that P(A È B)=3/4,

(a) equally likely and mutually exclusive

P(A Ç B)=1/4, P( A ) =2/3 then P ( A Ç B) is [2002]

(b) equally likely but not independent

(a) 5/12

(b) 3/8

(d) mutually exclusive and independent

(c) 5/8

(d) 1/4 3x + 1 1- x , P( B) = and 3 4

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(AÈB) is [2008] 3 5

1 - 2x P (C ) = The set of possible values of x 2

(a)

are in the interval.

(c) 1

(a) [0 , 1] é1 2 ù

(c) ê 3 , 3 ú û ë

4.

5.

Events A, B, C are mutually exclusive events such that P ( A) =

3.

(c) independent but not equally likely

[2003] é1 1 ù

(b) ê 3 , 2 ú û ë é 1 13 ù

(d) ê 3 , 3 ú û ë

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is [2003] (a)

2 5

(b)

4 5

(c)

3 5

(d)

1 5

Let A and B be two events such that 1 1 1 P ( A È B ) = , P ( A Ç B ) = and P ( A) = , 4 6 4

where A stands for complement of event A. Then events A and B are [2005]

6.

(b) 0

2 5 Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ...20}. [2010]

(d)

Statement -1: The probability that the chosen numbers when arranged in some order will form an AP is

1 . 85

Statement -2 : If the four chosen numbers form an AP, then the set of all possible values of common difference is (±1, ±2, ±3, ±4, ±5) . (a) Statement -1 is true, Statement -2 is true ; Statement -2 is not a correct explanation for Statement -1 (b) Statement -1 is true, Statment -2 is false (c) Statement -1 is false, Statment -2 is true. (d) Statement -1 is true, Statement -2 is true ; Statement -2 is a correct explanation for Statement -1.

M-87

Probability

7.

8.

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is : [2012] (a) 880 (b) 629 (c) 630 (d) 879 For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs)

1 and 4 1 P(All the three events occur simultaneously) = . 16 Then the probability that at least one of the events occurs, is : [2017] 3 7 (a) (b) 16 32 7 7 (c) (d) 16 64

= P(Exactly one of C or A occurs) =

Answer Key 1 (a)

1.

2 (b)

3 (a)

4 (c)

5 (c)

6 (b)

7 (d)

8 (c)

P ( A È B È C ) = P ( A) + P ( B ) + P ( C )

(a) P (A È B) = P (A) + P (B) – P (A Ç B);

ÞP ( A È B È C ) =

3 1 Þ =1 – P( A ) + P(B) – 4 4

Þ 1=1–

1 + 3x 1 - x 1 - 2 x + + £1 3 4 2 0 £ 13 - 3x £ 12 Þ 1 £ 3 x £ 13

2 2 + P(B) Þ P(B) = ; 3 3

\ 0£

Now, P( A Ç B ) = P(B) – P ( A Ç B ) = 2.

(b)

1 13 £x£ 3 3 Considering all inequations, we get Þ

5 2 1 – = . 3 4 12

3x + 1 1- x , P( B) = , 3 4 1 - 2x P (C ) = 2 P ( A) =

Q For any event E , 0 £ P ( E ) £ 1

Þ 0 £ 3 x + 1 £ 1, 0 £ 1 - x £ 1 3 4 1 - 2x and 0 £ £1 2 Þ -1 £ 3x £ 2, - 3 £ x £ 1 and - 1 £ 2 x £ 1 1 2 Þ - £ x £ £ -3 £ x £ 1, and 3 3

1 1 £x£ 2 2 Also for mutually exclusive events A, B, C, -

3x + 1 1 - x 1 - 2 x + + 3 4 2

1 1ü ì 1 ì 2 1 13 ü maxí- ,-3,- , ý £ x £ min í ,1, , ý 2 3þ î 3 î3 2 3 þ 1 1 é1 1 ù £ x £ Þ xÎê , ú 3 2 ë3 2 û

3.

(a) Let 5 horses are H1, H2, H3, H4 and H5. Selected pair of horses will be one of the 10 pairs (i.e.; 5

C2 ): H1 H2, H1 H3, H1 H4, H1 H5, H2H3, H2

H4, H2 H5, H3 H4, H3 H5 and H4 H5. Any horse can win the race in 4 ways. For example : Horses H2 win the race in 4 ways H1 H2, H2H3, H2H4 and H2H5. Hence required probability =

4 2 = 10 5

EBD_7764 M-88

4.

(c)

Mathematics

P ( A È B) = P ( A) =

1 1 , P ( A Ç B ) = and 6 4

Prob. =

1 4

5 3 Þ P ( A È B ) = , P( A) = 6 4 Also Þ P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B )

Þ P( B) =

7.

5 3 1 1 - + = 6 4 4 3

3 1 1 - = = P( A Ç B) 4 3 4 Hence A and B are independent but not equally likely. (c) A º number is greater than 3 3 1 Þ P( A) = = 6 2 B º number is less than 5 Þ P ( A) P ( B ) =

5.

4 2 Þ P( B) = = 6 3 A Ç B º number is greater than 3 but less than 5.

1 6 \ P(A È B) = P(A) + P(B) – P(A Ç B)

Þ P( A Ç B) =

= 6.

1 2 1 3 + 4 –1 + – = =1 2 3 6 6

(b) n(S) = 20C4 Statement-1: common difference cases = 17 common difference cases = 14 common difference cases = 11 common difference cases = 8 common difference cases = 5 common difference cases = 2

is 1; total number of

17 + 14 + 11 + 8 + 5 + 2 20

C4

1 85

Statement -2 is false, because common difference can be 6 also. (d) Number of white balls = 10 Number of green balls = 9 and Number of black balls = 7 \ Required probability = (10 + 1) (9 + 1) (7 + 1) – 1 = 11.10.8 –1 = 879 [Q The total number of ways of selecting one or more items from p identical items of one kind, q identical items of second kind; r identical items of third kind is (p + 1) (q + 1) (r + 1) –1 ] 64 127 P (exactly one of A or B occurs) =

8.

(c)

1 4 P (Exactly one of B or C occurs)

= P(A) + P (B) – 2P (A Ç B) =

1 4 P (Exactly one of C or A occurs)

= P(B) + P (C) – 2P (B Ç C) =

= P(C) + P(A) – 2P (C Ç A) =

1 4

...(1)

...(2)

...(3)

Adding (1), (2) and (3),we get 2SP(A) – 2SP (A Ç B) =

3 4

\ SP(A) – SP (A Ç B) =

3 8

is 2; total number of is 3; total number of

=

Now, P (A Ç B Ç C) =

1 16

is 4; total number of

\ P (A È B È C) = SP (A) – SP (A Ç B) + P (A Ç B Ç C)

is 5; total number of

=

is 6; total number of

3 1 7 + = 8 16 16

Relations and Functions 1.

2.

The period of sin 2 q is

interval of S is (a) [ –1, 3] (c) [ 0, 1]

[2002]

(b) (a) p 2 (c) 2 p (d) Which one is not periodic? (a) | sin3x | +sin 2x

p p /2

[2002]

6.

(b) cos x + (c) cos 4x + tan2x (d) cos2x + sinx

4.

(

2

)

Let R = {(1,3), (4, 2), (2, 4), (2, 3), (3,1)} be a

neither one -one nor onto one-one but not onto onto but not one-one one-one and onto both.

If f : R ® S , defined by

f ( x) = sin x - 3 cos x + 1, is onto, then the

R is [2004] (a) reflexive (b) transitive (c) not symmetric (d) a function Let f : (– 1, 1) ® B, be a function defined by -1 f (x) = tan

[2003] (a) neither an even nor an odd function (b) an even function (c) an odd function (d) a periodic function. A function f from the set of natural numbers to integers defined by [2003] ìn -1 ïï 2 , when n is odd is = f (n) í ï - n , when n is even ïî 2

5.

7.

The function f ( x) = log x + x + 1 , is

(a) (b) (c) (d)

[2004] (b) [–1, 1] (d) [0, 3]

relation on the set A = {1, 2,3, 4}. . The relation

cos2x

3.

16

2x 1 - x2

, then f is both one - one and

onto when B is the interval

8.

9.

[2005]

(a)

æ pö ç 0, ÷ è 2ø

(b)

é pö ê0, ÷ ë 2ø

(c)

é p pù ê- 2 , 2 ú û ë

(d)

æ p pö ç- , ÷ è 2 2ø

A real valued function f (x) satisfies the functional equation f (x – y) = f (x) f (y) – f (a – x) f (a + y) where a is a given constant and f (0) = 1, f (2a – x) is equal to [2005] (a) – f (x) (b) f (x) (c) f (a) + f (a – x) (d) f (– x) Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. The relation is [2005] (a) reflexive and transitive only (b) reflexive only (c) an equivalence relation (d) reflexive and symmetric only

EBD_7764 M-90 10. Let W denote the words in the English dictionary. Define the relation R by R = {(x, y) Î W × W| the words x and y have at least one letter in common.} Then R is [2006] (a) not reflexive, symmetric and transitive (b) relexive, symmetric and not transitive (c) reflexive, symmetric and transitive (d) reflexive, not symmetric and transitive 11. Let f: N®Y be a function defined as f(x) = 4x + 3 where Y = {y Î N : y = 4x + 3 for some x Î N}. Show that f is invertible and its inverse is [2008]

(a)

13.

3y + 4 3

(b) g ( y ) = 4 +

15.

y+3 4

y+3 y –3 (d) g ( y ) = 4 4 Let R be the real line. Consider the following subsets of the plane R × R: S ={(x, y): y = x + 1 and 0 < x < 2} T ={(x, y): x – y is an integer}, Which one of the following is true? [2008] (a) Neither S nor T is an equivalence relation on R (b) Both S and T are equivalence relation on R (c) S is an equivalence relation on R but T is not (d) T is an equivalence relation on R but S is not DIRECTIONS : This question contains two statements: Statement-1 (Assertion ) and Statement-2 (Reason). This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

(c)

12.

g ( y) =

14.

g ( y) =

2

16.

17.

Mathematics Statement-2 is not a correct explanation for Statement-1. For real x, let f (x) = x3 + 5x + 1, then [2009] (a) f is onto R but not one-one (b) f is one-one and onto R (c) f is neither one-one nor onto R (d) f is one-one but not onto R Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w};

æ m pö S = {ç , ÷ | m,n, p and q are integers such è n qø that n, q ¹ 0 and qm = pn}. Then [2010] (a) Neither R nor S is an equivalence relation (b) S is an equivalence relation but R is not an equivalence relation (c) R and S both are equivalence relations (d) R is an equivalence relation but S is not an equivalence relation Let R be the set of real numbers. [2011] Statement-1: A = {(x, y) Î R × R : y – x is an integer} is an equivalence relation on R. Statement-2: B = {(x, y) Î R × R : x = ay for some rational number a} is an equivalence relation on R. (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Let f be a function defined by

Let f(x) = ( x + 1) –1, x ³ –1

f ( x) = ( x -1) +1, ( x ³1) .

Statement -1 : The set {x : f(x) = f –1(x) = {0, –1}

Statement - 1 :

Statement-2 : f is a bijection. [2009] (a) Statement-1 is true, Statement-2 is true. Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true.

2

{

[2011RS]

}

-1 The set x : f ( x) = f ( x) = {1, 2} .

Statement - 2: f is a bijection and f -1 ( x ) = 1 + x - 1, x ³ 1. (a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

M-91

Relations and Functions

(b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. 18.

(d) Statement-1 is false, Statement-2 is true. If g is the inverse of a function f and

f '( x ) = (a)

1 1 + x5

, then g ¢ ( x ) is equal to:[2014]

1

(c) 1 + x5

(d) 5x4

é 1 1ù 19. The function f : R ® ê - , ú defined as f(x) = ë 2 2û x , is : [2017] 1 + x2 (a) neither injective nor surjective (b) invertible (c) injective but not surjective (d) surjective but not injective

(b) 1 + { g ( x )}

5

1 + { g ( x )}

5

Answer Key 1 (b) 16 (a)

2 (b) 17 (a)

3 (c) 18 (b)

4 (d) 19 (d)

5 (a)

6 (c)

7 (d)

8 (a)

9 (a)

(b)

2.

(b) Q cos x is non periodic \ cos x + cos 2 x can not be periodic.

3.

(c)

5.

f ( x ) = log( x + x 2 + 1)

{

}

f( - x ) = log - x + x 2 + 1 ìï - x 2 + x 2 + 1üï = log í ý ïî x + x 2 + 1 ïþ

= - log( x + x 2 + 1) = - f ( x ) Þ f(x) is an odd function. 4.

(d) We have f : N ® I If x and y are two even natural numbers, -x - y = Þx=y 2 2 Again if x and y are two odd natural numbers then

then f ( x) = f ( y ) Þ

f ( x) = f ( y) Þ

x -1 y - 1 = Þx=y 2 2

11 (d)

12 (d)

13 (b)

14 (b)

15 (b)

\ f is onto. Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number. \ f is onto. Hence f is one one and onto both.

1 - cos 2q 2p sin q = =p ; Period = 2 2 2

1.

10 (b)

(a)

f ( x) is onto

\ S = range of f (x)

Now f (x) = sin x - 3cos x + 1 pö æ = 2sin ç x - ÷ + 1 è 3ø

pö æ Q -1 £ sin ç x - ÷ £ 1 è 3ø pö æ -1 £ 2sin ç x - ÷ + 1 £ 3 è 3ø \ f ( x) Î[ -1, 3] = S We know that - a 2 + b2 £ a sin q + b cos q £ a 2 + b2 \ -2 £ sin x - 3 cos x £ 2

EBD_7764 M-92

Mathematics

Also f (x) = 4x + 3 = y

Þ -1 £ sin x - 3 cos x + 1 £ 3

\ f ( x) Î[ -1, 3] 6.

7.

(c) Q (1, 1) Ï R Þ R is not reflexive (2,3) Î R but (3, 2) Ï R \ R is not symmetric æ 2x ö (d) Given f (x) = tan -1 ç = 2tan–1x è 1 - x 2 ÷ø

for x Î (-1, 1)

æ -p , If x Î( -1, 1) Þ tan -1 x Î ç è 4

pö ÷ 4ø

æ -p p ö Þ 2 tan -1 x Î ç , è 2 2 ÷ø æ p pö Clearly, range of f (x) = ç - , ÷ è 2 2ø For f to be onto, codomain = range \ Co-domain of function = B =

8.

æ p pö ç- , ÷ . è 2 2ø (a) f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (0) f (x) = f (a) f (x –a) – f (x) = – f (x)

[Q x = 0, y = 0, f (0) = f 2 (0) - f 2 (a) 2

Þ f (a) = 0 Þ f (a) = 0

9.

y –3 y –3 \ g ( y) = 4 4 12. (d) Given S = {(x , y) : y = x + 1 and 0 < x < 2} Q x ¹ x + 1 for any x Î(0, 2) Þ (x, x) Ï S \ S is not reflexive. Hence S in not an equivalence relation. Also T ={x, y): x – y is an integer} Q x – x = 0 is an integer " x Î R \ T is reflexive. If x – y is an integer then y – x is also an integer \T is symmetric If x – y is an integer and y – z is an integer then (x – y) + (y– z) = x – z is also an integer. \ T is transitive 13. (b) Given that f (x) = (x + 1)2 –1, x ³ –1 Clearly Df = [–1, ¥ ) but co-domain is not given. Therefore f (x) need not be necessarily onto. But if f (x) is onto then as f (x) is one one also, (x + 1) being something +ve, f –1(x) will exist where (x + 1)2 –1 = y

Þx=

Þ f (2a - x) = - f ( x ) (a) Reflexive and transitive only. e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive] (3, 6), (6, 12), (3, 12) [Transitive]. (3, 6) Î R but (6, 3) Ï R [ non symmetric]

10.

(b) Clearly ( x, x) Î R"x ÎW. So R is reflexive.

11.

Let ( x, y) Î R , then ( y, x) Î R as x and y have at least one letter in common. So, R is symmetric. But R is not transitive for example Let x = INDIA, y = BOMBAY and z =JOKER then ( x, y ) Î R (A is common) and ( y, z ) Î R (O is common) but ( x, z ) Ï R . (as no letter is common) (d) Clearly f is one one and onto, so invertible

Þ x +1 =

y +1

(+ve square root as x +1 ³ 0 ) Þ x =–1+

y+1

Þ f –1 (x) = x + 1 – 1 Then f (x) = f –1 (x) Þ (x + 1)2 – 1 = x + 1 –1 Þ (x + 1)2 = x + 1 Þ (x + 1)4 = (x + 1) Þ (x + 1) [ (x + 1)3 – 1] = 0 Þ x = – 1, 0

\ The statement-1 is correct but statement-2 is false. 14. (b) Given that f (x) = x3 + 5x + 1 \f ' (x) = 3x2 + 5 > 0, "xÎR Þ f (x) is strictly increasing on R Þ f (x) is one one \ Being a polynomial f (x) is continuous and increasing. on R with lim f ( x ) = -¥ x ®¥

f ( x) = ¥ and xlim ®¥

Relations and Functions

15.

\ Range of f = ( - ¥, ¥) = R Hence f is onto also. So, f is one one and onto R. (b) x Ry need not implies yRx m p S: s n q p m Given qm = pn Þ = q n m p p m m m \ s reflexive n s q Þ q s n symmetric n n m p p r s , s qm = pn, ps = rq n q q s Þ

p m r Þ q = n = s Þ ms = rn transitive. S is an equivalence relation. 16. (a) Let for statement 1: xRy = x – y Î I . As xRx is an integer and yRx as well as xRz (for xRy and yRz) is also an integer. Hence equivalence. Similarly as x = ay hence a =1 for reflexive 1 and being a rational for symmetric for a some non zero a and product of rationals also being rational Þ equivalence But not symmetric because of a = 0 case Both relations are equivalence but not the correct explanation. 17.

(a)

f ( x ) = ( x - 1) + 1, x ³ 1

M-93 Hence statement-1 is correct 18. (b) Since f (x) and g(x) are inverse of each other

g'( f (x)) =

Þ

g '( f ( x)) = 1 + x5

1 ö æ çèQ f ¢ ( x) = ÷ 1 + x5 ø

Here x = g(y) g ¢( y ) = 1 + [ g ( y )]

5

\

Þ g ¢ ( x ) = 1 + ( g ( x) ) 5 19. (d)

é 1 1ù we have f : R ® ê - , ú , ë 2 2û x "x Î R f (x) = 1 + x2

Þ f ¢(x) =

Þ

y = ( x - 1) + 1 Þ ( x - 1) = y - 1

Þ

2

y - 1 Þ f -1 ( y ) = 1 ± y - 1

f -1 ( x ) = 1 + x - 1 {\ x ³ 1}

Hence statement-2 is correct Now f ( x ) = f

-1

( x)

Þ f ( x ) = x Þ ( x - 1) + 1 = x 2

Þ

(1 + x )

x = –1

=

x 2 - 3 x + 2 = 0 Þ x = 1, 2

-(x + 1)(x - 1) (1 + x 2 ) 2

– x=1

sign of f¢ (x) Þ f¢ (x) changes sign in different intervals. \ Not injective

x 1 + x2

Þ y + yx2 = x Þ yx2 – x + y = 0

f :[1, ¥) ® [1, ¥)

Þ x = 1±

2 2

+

Now y =

\

2

(1 + x 2 ).1 - x.2x

–

2

Since f is a bijective function

1 f '( x )

\

For y ¹ 0, D = 1 – 4y2 ³ 0

é -1 1 ù Þ y Î ê , ú - {0} ë 2 2û For y = 0 Þ x = 0 é -1 1 ù \ Range is ê , ú ë 2 2û Þ Surjective but not injective

EBD_7764 M-94

Mathematics

Inverse Trigonometric Functions 1.

cot -1 ( cos a ) - tan -1 ( cos a ) = x , then sin x =

(a)

æaö tan2 ç ÷ è2ø

3.

4.

The domain (a) [1, 9] (c) [–9, 1]

of sin-1 [log

9- x

2

(a) [1, 2] (c) [1, 2 ]

7.

9.

x
0 Þ -3 < x < 3 Taking common solution of (i) and (ii), we get 2 £ x < 3 \ Domain = [2, 3) y (c) cos - 1 x - cos - 1 = a 2 æ ö æ xy y2 ö cos - 1 ç + (1 - x 2 ) ç1 - ÷ ÷ = a 4 ø÷ çè 2 è ø

= (1– cosa) and base = 2 cos a ]

2.

1

if (i) - 1 £ x - 3 £ 1 Þ 2 £ x £ 4

cos a 1 - cos a =x 2 cos a

Þ tan x =

(b)

- cos a . cos a

10 (b)

\a £

( cos a ) = x

ö ÷÷ – tan–1 ( cos a ) = x ø 1

tan–1

5 (c)

Þ

æ xö p æ 5ö sin -1 ç ÷ = - cosec -1 ç ÷ è 5ø 2 è 4ø æ xö p æ 4ö sin -1 ç ÷ = - sin -1 ç ÷ è 5ø 2 è 5ø

[Q sin -1 x + cos -1 x = p / 2] Þ

æ xö æ 4ö sin -1 ç ÷ = cos -1 ç ÷ è 5ø è 5ø

Let cos -1

4 4 = A Þ cos A = 5 5

....(i)

EBD_7764 M-96

5 A

Þ

Þ

é –1 17 ù = cot ê tan 6 úû ë æ –1 6 ö 6 = cot çè cot ÷= 17 ø 17

3 B

4

A = cos–1 (4/5)

Þ sin A =

9.

3 5

x 3 = sin -1 5 5

Þ

3

æx ö f (x) is defined if – 1 £ ç - 1÷ £ 1 and è2 ø cos x > 0 or

x p p 0 £ £ 2 and - < x < 2 2 2

or

0 £ x £ 4 and -

\

é pö x Î ê 0, ÷ ë 2ø

p p < x< 2 2

5 2ö æ cot ç cos ec –1 + tan –1 ÷ = 3 3ø è 3 2ù é cot ê tan –1 + tan –1 ú 4 3û ë é æ 3 2 öù ê –1 ç 4 + 3 ÷ ú ú = cot ê tan ç 3 2÷ ê ç1– ´ ÷ ú è êë 4 3 ø úû

3ù é -1 3x - x tan–1 y = tan ê 2ú ëê 1 - 3x ûú

x 3 = Þ x= 5 5

2 æx ö (b) f (x) = 4- x + cos -1 ç - 1÷ + log(cos x) è2 ø

(a)

= tan -1 x + 2 tan -1 x = 3 tan–1x

3 5 –1 cos (4/5) = sin–1 (3/5) equation (i) become,

sin -1

8.

é 2x ù (c) tan–1 y = tan -1 x + tan -1 ê ú ë1 - x 2 û

A = sin –1

\ \

7.

Mathematics

C

Þ y=

10. (b)

3x - x 3 1 - 3x 2

æ 6x x ö Let F(x) = tan –1 çç 3 ÷÷ where x è 1 - 9x ø

æ 1ö Î ç 0, ÷ . è 4ø æ 2.(3x3 / 2 ) = tan–1 çç 3/ 2 2 è 1 - (3x )

ö ÷ = 2 tan–1 (3x3/2) ÷ ø

æ 3ö As 3x3/2 Î ç 0, ÷ è 8ø 1 3ù é 3/ 2 1 < Þ 0 < 3x 3/ 2 < ú êQ 0 < x < 4 Þ 0 < x 8 8û ë So

1 dF(x) 3 × 3 × × x1/2 =2× 3 dx 2 1 + 9x 9

x 1 + 9x 3 On comparing =

\ g(x) =

9 1 + 9x 3

M-97

Matrices

10-13 delet

Matrices 1.

éa b ù 2 éa b ù If A = ê ú and A = ê b a ú , then b a ë û ë û

[2003] 2

(a) a = 2 ab, b = a + b 2

2

5.

2

(b) a = a + b , b = ab (c) a = a 2 + b 2 , b = 2ab (d) a = a 2 + b 2 , b = a 2 - b 2 . 2.

é1 0ù If A = ê ú and I = ë1 1 û

é1 0 ù ê ú , then which one of ë0 1û

6.

the following holds for all n ³ 1, by the principle of mathematical induction [2005]

3.

(a)

An = nA – (n – 1) I

(b)

An = 2n - 1 A – (n – 1) I

(c)

An = nA + (n – 1) I

n (d) A = 2n - 1 A + (n – 1) I If A and B are square matrices of size n × n such

that A2 - B 2 = ( A - B)( A + B) , then which of the following will be always true? [2006] (a) A = B (b) AB = BA (c) either of A or B is a zero matrix (d) either of A or B is identity matrix 4.

æ 1 2ö æ a 0ö Let A = ç and B = ç , a, b Î N. ÷ è 3 4ø è 0 b÷ø

Then [2006] (a) there cannot exist any B such that AB = BA (b) there exist more than one but finite number

7.

18

of B¢s such that AB = BA (c) there exists exactly one B such that AB = BA (d) there exist infinitely many B¢s such that AB = BA The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is [2010] (a) 5 (b) 6 (c) at least 7 (d) less than 4 Let A and B be two symmetric matrices of order 3. [2011] Statement-1: A(BA) and (AB)A are symmetric matrices. Statement-2: AB is symmetric matrix if matrix multiplication of A with B is commutative. (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. If w ¹ 1is the complex cube root of unity and

éω matrix H = ê ë0 (a) 0 (c) H2

0ù ú , then H70 is equal to ωû [2011RS] (b) –H (d) H

EBD_7764 M -9 8

8.

Mathematics

+ b is equal to : (a) 4 (b) 13

é1 2 2 ù ê ú If A = ê 2 1 -2ú is a matrix satisfying the êë a 2 b úû

equation AAT = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to: [2015] (a) (2, 1) (b) (–2, – 1) (c) (2, – 1) (d) (–2, 1) 9.

[2016] (d) 5

(c) –1

é 2 -3ù 2 10. If A = ê ú , then adj (3A + 12A) is equal to ë -4 1 û : [2017]

é5a - b ù ú and A adj A = A AT, then 5a If A = ê ê ë 3 2 úû

é 72 -63ù (a) ê ú ë -84 51 û

é 72 -84 ù (b) ê ú ë -63 51 û

é 51 63ù (c) ê ú ë84 72 û

é 51 84 ù (d) ê ú ë 63 72 û

Answer Key 1 (c)

1.

(c)

2 (a)

3 (b)

4 (d)

5 (c)

6 (a)

7 (d)

8 (b)

éa b ù é a b ù é a b ù A2 = ê ú=ê úê ú ë b a û ë b a û ëb a û éa 2 + b 2 =ê êë 2ab

2ab ù ú a 2 + b2 úû

a = a 2 + b 2 ; b = 2ab

2.

9 (d)

3.

(b)

Þ AB = BA 4.

(d)

é1 A= ê ë3

2ù 4 úû

éa B = ê ë0

0ù b úû

é a 2b ù AB = ê ú ë3a 4b û

é 1 0 ù 3 é1 0 ù A2 = ê ú , A = ê3 1 ú and we can ë2 1 û ë û

é a 0 ù é1 2 ù é a 2 a ù BA = ê úê ú=ê ú ë 0 b û ë3 4û ë3b 4b û

é 1 0ù prove by induction that An = ê ú ë n 1û

Hence, AB = BA only when a = b \ There can be infinitely many B¢s for which AB = BA

0 ù é n 0ù é n - 1 Now nA - ( n - 1) I = ê n n ú - ê 0 n - 1úû ë û ë

\ nA - ( n - 1) I = An

A2 - B 2 = ( A - B )( A + B )

A2 - B 2 = A2 + AB - BA - B 2

(a) We observe that

é 1 0ù n = ê n 1ú = A ë û

10 (c)

5.

(c)

é 1 ... ...ù ê... 1 ...ú ê ú are 6 non-singular matrices êë... ... 1 úû

M-99

Matrices

because 6 blanks will be filled by 5 zeros and 1 one.

é 1+ 4 + 4 2+ 2- 4 a + 4 + 2b ù é9 0 0ù ê ú 4 +1+ 4 2a + 2 - 2b ú = ê0 9 0ú Þê 2 + 2 - 4 ê ú ê 2 2 ú ê0 0 9 ú û ëa + 4 + 2b 2a + 2 - 2b a + 4 + b û ë

é... ... 1 ù Similarly, ê... 1 ...ú are 6 non-singular ê ú êë 1 ... ...úû

6.

7.

matrices. So, required cases are more than 7, nonsingular 3 × 3 matrices. (a) \ A¢ = A B¢ = B Now (A(BA))¢ = (BA)¢A¢ = (A¢B¢)A¢ = (AB)A = A(BA) Similarly ((AB)A)¢ = (AB)A So, A(BA) and (AB)A are symmetric matrices. Again (AB)¢ = B¢A¢ = BA Now if BA = AB, then AB is symmetric matrix. (d)

Þ a + 4 + 2b = 0 Þ a + 2b = – 4 ...(i) 2a + 2 – 2b = 0 Þ 2a – 2b = – 2 Þ a – b = –1 On solving (i) and (ii) we get – 1 + b + 2b =

1

and a = – 2 (a, b) = (–2, –1) 9.

(b)

A(adj A) = A AT Þ A–1A (adj A) = A–1A AT adj A = AT

é 2 b ù é5a 3 ù ú=ê ú Þê êë-3 5a úû êë- b 2 úû 2 Þ a = and b = 3 5

10. (c)

éωk +1 0 ù ú =ê êë0 ωk +1 úû

é 2 -3ù We have A = ê -4 1 ú ë û é 16 -9 ù Þ A2 = ê ú ë -12 13 û

So by principle of mathematical induction,

8.

(d)

Þ 5a + b = 5

éωk 0ù H =ê ú then Hk + êë0 ω úû

éω H 70 = ê êë0

...(i)

3b = – 3 b = –1

k

70

–4

– 1 + 3b = – 4

2 éω 0 ù é ω 0 ù éω 0 ù ê ú H2 = ê = úê ú ω2 úû ë0 ωû ë 0 ωû êë0

If

...(ii)

0 ù éω w 0 ù éω ú =ê ú =ê ω70 úû êë0 w 69ωúû ë0 69

é 1 2 2 ù é 1 2 a ù é 9 0 0ù ê 2 1 -2ú ê2 1 2ú = ê0 9 0ú ê úê ú ê ú ëê a 2 b ûú êë2 -2 b ûú ëê0 0 9 ûú

0ù ú=H ωû

é 48 -27 ù Þ 3A2 = ê -36 39 ú ë û é 24 -36 ù Also 12A = ê ú ë -48 12 û \ 3A2 + 12A é 48

= ê -36 ë

-27 ù é 24 -36 ù é 72 -63ù + 39 úû êë -48 12 úû = êë -84 51 úû

é 51 63ù adj (3A2 + 12A) = ê ú ë84 72 û

EBD_7764 M-100

Mathematics

10-13 delet

19

Determinants 1.

If a > 0 and discriminant of ax2+2bx+c is –ve, (a)

A2 = I (b) A = (–1) I, where I is a unit matrix

ax + b bx + c is equal to then ax + b bx + c 0 a b

b c

-1 (c) A does not exist (d) A is a zero matrix

[2002]

2.

(a) +ve (b) (ac-b2)(ax2+2bx+c) (c) –ve (d) 0 If the system of linear equations

5.

If B is the inverse of matrix A, then a is (a) 5 (b) –1 (c) 2 (d) –2

[2003]

x + 2ay + az = 0 ; x + 3by + bz = 0 ;

x + 4cy + cz = 0 has a non - zero solution, then a, b, c. (a) satisfy a + 2b + 3c = 0 (c) are in G..P 3.

wn

w 2n

D = wn

w2n

1

w

(a) w 2 (c) 1

4.

6.

(b) are in A.P (d) are in H.P.

If 1, w, w 2 are the cube roots of unity, then 1 2n

1

w

is equal to

[2003]

n

7.

(b) 0 (d) w

æ 0 0 -1ö Let A = ç 0 -1 0 ÷ . The only correct ç ÷ è -1 0 0 ø

statement about the matrix A is

[2004]

æ 1 -1 1 ö æ 4 2 2ö ç ÷ Let A = 2 1 -3 . and B = ç -5 0 a ÷ . ç ÷ ç ÷ è1 1 1 ø è 1 -2 3 ø

8.

[2004]

If a1, a2 , a3 ,......, an ,.... are in G.P., then the value of the determinant [2004] log an

log an+1

log an+ 2

log an+ 3

log an+ 4

log an +5

log an+ 6

log an+ 7

log an+8

(a) –2 (c) 2 The system of equations ax+y +z= a –1 x + a y+ z = a – 1 x+ y+ az = a –1 has infinite solutions, if a (a) – 2 (c) not – 2

, is

(b) 1 (d) 0

is [2005] (b) either – 2 or 1 (d) 1

2 2 2 If a + b + c = – 2 and

1 + a2 x

[2005]

(1 + b 2 ) x (1 + c 2 ) x

2 2 2 f (x) = (1 + a ) x 1 + b x (1 + c ) x , (1 + a2 ) x (1 + b 2 ) x 1 + c 2 x

M-101

Determinants

9.

then f (x) is a polynomial of degree (a) 1 (b) 0 (c) 3 (d) 2 If a1 , a2 , a3 , ............, an , ...... are in G. P., then the determinant log an

D = log an + 3 log an + 6

log an + 1

log an + 2

log an + 4

log an + 5

log an + 7

log an + 8

is equal to (a) 1 (c) 4 10.

2 If A – A + I = 0 , then the inverse of A is [2005] (a) A + I (b) A (c) A – I (d) I – A

1

11.

13.

1

1

1 for x ¹ 0, y ¹ 0 , then D If D = 1 1 + x 1 1 1+ y

is (a) (b) (c) (d) 12.

[2005] (b) 0 (d) 2

[2007] divisible by x but not y divisible by y but not x divisible by neither x nor y divisible by both x and y

5 5a a 2 Let A = 0 a 5a . If A = 25 , then a 0 0 5 equals [2007] (a) 1/5 (b) 5 (c) 52 (d) 1 Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A), the sum of diagonal entries of a. Assume that A2 = I. [2008]

for Statement-1 (d) Statement -1 is true, Statement-2 is false 14. Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to [2008] (a) 2

(b) –1

(c) 0

(d) 1

15. Let A be a square matrix all of whose entries are integers. Then which one of the following is true? [2008] (a) If det A = ± 1, then A–1 exists but all its entries are not necessarily integers (b) If det A ¹ ± 1, then A–1 exists and all its entries are non integers (c) If det A = ± 1, then A–1 exists but all its entries are integers (d) If det A = ± 1, then A–1 need not exists 16. Let A be a 2 × 2 matrix Statement -1 : adj (adj A) = A Statement -2 : |adj A |= |A|

(a) Statement-1 is true, Statement-2 is true. Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement -1 is false, Statement-2 is true. (d) Statement-1 is true, Statement -2 is true. Statement-2 is a correct explanation for Statement-1. 17. Let a, b, c be such that b(a + c) ¹ 0 if [2009] a

a + 1 a –1

Statement-1 : If A ¹ I and A ¹ –I, then det (A) = –1

–b b + 1 b –1

Statement-2 : If A ¹ I and A ¹ –I, then tr (A) ¹ 0.

c

(a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation

[2009]

c –1 c + 1

+

a +1

b +1

c –1

a –1

b -1

c + 1 = 0,

(-1)n+ 2 a (-1)n +1b (-1)n c

then the value of n is : (a) any even integer

EBD_7764 M-102

18.

19.

20.

21.

(b) any odd integer (c) any integer (d) zero Let A be a 2 × 2 matrix with non-zero entries and let A2 = I , where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement - 1 : Tr(A) = 0. Statement -2 : |A| = 1. [2010] (a) Statement -1 is true, Statement -2 is true ; Statement -2 is not a correct explanation for Statement -1. (b) Statement -1 is true, Statement -2 is false. (c) Statement -1 is false, Statement -2 is true . (d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1. Consider the system of linear equations; x1 + 2x2 + x3 = 3 [2010] 2x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1 The system has (a) exactly 3 solutions (b) a unique solution (c) no solution (d) infinite number of solutions The number of values of k for which the linear equations 4x + ky + 2z = 0 , kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is [2011] (a) 2 (b) 1 (c) zero (d) 3 If the trivial solution is the only solution of the system of equations [2011RS] x - ky + z = 0

–1 R = { ( A, B) A = P BP for some invertible matrix P} Statement-1 : R is equivalence relation. Statement-2 : For any two invertible 3 ´ 3

matrices M and N, ( MN )

3x + y - z = 0 then the set of all values of k is : (a)

R - { 2, -3}

(b)

R - { 2}

(c)

R - { -3}

(d)

{ 2, -3}

Statement - 1: Determinant of a skew-symmetric matrix of order 3 is zero.

-1

= N -1 M -1 .

(a) Statement-1 is true, statement-2 is true and statement-2 is a correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1. (c) Statement-1 is true, stement-2 is false. (d) Statement-1 is false, statement-2 is true. 24.

æ1 0 0ö Let A = çç 2 1 0 ÷÷ . If u1 and u2 are column ç 3 2 1÷ è ø æ0ö æ1ö ç ÷ ç ÷ matrices such that Au1 = ç 0 ÷ and Au2 = ç 1 ÷ , ç0÷ ç0÷ è ø è ø then u1 + u2 is equal to : [2012]

(a)

kx + 3 y - kz = 0

22

Mathematics Statement - 2 : For any matrix A, det (A)T= det (A) and det (– A) = – det (A). Where det (B) denotes the determinant of matrix B. Then : [2011RS] (a) Both statements are true (b) Both statements are false (c) Statement-1 is false and statement-2 is true (d) Statement-1 is true and statement-2 is false 23. Consider the following relation R on the set of real square matrices of order 3. [2011RS]

æ -1 ö ç ÷ ç 1÷ ç0÷ è ø

(b)

æ -1 ö ç ÷ ç 1÷ ç -1 ÷ è ø

æ1ö æ -1 ö ç ÷ ç ÷ (c) ç -1 ÷ (d) ç -1 ÷ ç -1 ÷ ç0÷ è ø è ø 25. Let P and Q be 3 ´ 3 matrices P ¹ Q. If P3= Q3 and P2Q = Q2P then determinant of (P2 + Q2) is equal to : [2012] (a) – 2 (b) 1 (c) 0 (d) – 1

Determinants 26. The number of values of k, for which the system of equations : (k + 1) x + 8y = 4k kx + (k + 3)y = 3k – 1 has no solution, is [2013] (a) infinite (b) 1 (c) 2 (d) 3

27.

28.

é 1 a 3ù If P = ê 1 3 3ú is the adjoint of a 3 × 3 matrix ê ú êë 2 4 4úû A and |A| = 4, then a is equal to : [2013] (a) 4 (b) 11 (c) 5 (d) 0

If a, b ¹ 0, and f ( n ) = a n + b n and

(a)

1 + f (1) 1 + f ( 2 )

3

1 + f (1) 1 + f ( 2 ) 1 + f ( 3) 1 + f ( 2 ) 1 + f ( 3) 1 + f ( 4 )

= K (1 - a ) (1 - b ) ( a - b) 2

2

then K is equal to: (a) 1

29.

,

[2014]

(d)

(a) B –1 30.

2

(b) –1

ab

(b)

æ 1ö çè 2, ÷ø 2

1ö æ (b) ç 2, - ÷ è 2ø

3ö æ æ 3ö (d) ç1, - ÷ çè1, ÷ø è 4ø 4 33. If S is the set of distinct values of ‘b’ for which the following system of linear equations x + y+ z = 1 [2017] x + ay + z = 1 ax + by + z = 0 has no solution, then S is : (a) a singleton (b) an empty set (c) an infinite set (d) a finite set containing two or more elements 34. Let w be a complex number such that 2w + 1 = z (c)

1 ab If A is an 3 × 3 non-singular matrix such that AA' = A'A and B = A–1A', then BB' equals: [2014] (c)

M-103 (a) contains two elements. (b) contains more than two elements (c) is an empty set. (d) is a singleton 31. The system of linear equations [2016] x + ly – z = 0 lx – y – z = 0 x + y – lz = 0 has a non-trivial solution for: (a) exactly two values of l. (b) exactly three values of l. (c) infinitely many values of l. (d) exactly one value of l. 32. Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point : [2017]

( B )¢ -1

1

(c) I + B (d) I The set of all values of l for which the system of linear equations : [2015] 2x1 – 2x2 + x3 = lx1 2x1 – 3x2 + 2x3 = lx2 –x1 + 2x2 = lx3 has a non-trivial solution,

where z =

1

1

2 2 -3 . If 1 -w - 1 w = 3k, then

k is equal to : (a) 1 (c) z

1

w2

w7

(b) –z (d) –1

[2017]

Answer Key 1 (c) 16 (a) 31 (b)

2 (d) 17 (b) 32 (a)

3 (b) 18 (b) 33 (a)

4 (a) 19 (c) 34 (b)

5 (a) 20 (a)

6 (d) 21 (a)

7 (a) 22 (d)

8 (d) 23 (b)

9 (b) 24 (d)

10 (d) 25 (c)

11 (d) 26 (b)

12 (a) 27 (b)

13 (d) 28 (a)

14 (d) 29 (d)

15 (c) 30 (a)

EBD_7764 M-104

Mathematics

a

1.

(c) We have

ax + b

b

b

é 0 0 -1ù é 0 0 -1ù Also A = ê 0 -1 0 ú ê 0 -1 0 ú ê úê ú ëê -1 0 0 úû êë -1 0 0 ûú 2

bx + c

c

ax + b bx + c By R3 ® R3 – (xR1 + R2);

0

é1 0 0ù = ê0 1 0ú = I ê ú êë0 0 1úû

ax + b

a b

bx + c

= b c

2

2.

0 0 -(ax + 2bx + c) = (ax2 + 2bx + c)(b2 – ac) = (+)(–) = –ve. (d) For homogeneous system of equations to have non zero solution, D = 0 1 2a a

5.

é 4 2 2ù 1 ê ÞB= -5 0 a ú ú 10 ê êë 1 -2 3 úû

2

1 3b b = 0 C2 ® C2 - 2C3 b - 4ac 1 4c c 1 1

0 b

Also since, B = A-1 Þ AB = I

a b = 0 R3 ® R3 - R2 , R2 ® R2 - R1

é 1 -1 1 ù é 4 2 2 ù é1 0 0 ù 1 ê úê ú ê ú 2 1 -3ú ê -5 0 a ú = ê0 1 0 ú Þ 10 ê êë 1 1 1 úû êë 1 -2 3 úû êë0 0 1 úû

1 2c c

2 1 1 = + b a c \ a, b, c are in Harmonic Progression.

On simplification,

3.

(b)

1

wn

w 2n

D = wn

w 2n

1

1

wn

w 2n

(

)

(

)

=w

4.

-1- 0 + w

3n

é10 0 5 - 2 ù é1 0 0ù 1 ê ú ê ú 0 10 -5 + a ú = ê0 1 0ú Þ ê 10 êë 0 0 5 + a úû êë0 0 1 úû 5-a =0Þa =5 10 (d) Let r be the common ratio, then

Þ

(

3n n 2n 2n 2n n 4n = 1 w -1 - w w - w + w w - w 3n

é 4 2 2ù (a) Given that 10 B = ê -5 0 a ú ê ú êë 1 -2 3 úû

6n

-w é = 1 - 1 + 1 - 1 = 0 Q w3n = 1ù ë û é 0 0 -1ù ê ú (a) A = ê 0 -1 0 ú êë -1 0 0 úû clearly A ¹ 0. Also |A| = -1 ¹ 0

\ A-1 exists, further é -1 0 0 ù ( -1) I = ê 0 -1 0 ú ¹ A ê ú êë 0 0 -1úû

)

6.

log an

log an+1

log an+ 2

log an+ 3

log an+ 4

log an +5

log an+ 6

log an+ 7

log an+8

log a1r n -1

log a1r n

log a1r n +1

= log a1r n + 2

log a1r n + 3

log a1r n + 4

log a1r n +5

log a1r n + 6

log a1r n + 7

log a1 + ( n - 1) log r log a1 + n log r log a1 + (n + 1) log r = log a1 + ( n + 2) log r log a1 + (n + 3) log r log a1 + (n + 4) log r log a1 + ( n + 5) log r log a1 + (n + 6)log r log a2 + (n + 7) log r

M-105

Determinants

9.

1 1 ù é = 0 ê Apply c2 ® c2 - c1 - c3 ú 2 2 û ë 7.

(a)

(b) Q a1, a2 , a3 ,....... are in G.P.. \ Using an = ar n -1 ,we get the given determinant, as

ax + y + z = a - 1 x + a y + z = a – 1; x+ y+ z a = a –1

log ar n -1

log ar n

log ar n +1

a

1

1

log ar n + 2

log ar n + 3

log ar n + 4

D= 1

a

1

log ar n + 5

log ar n + 6

log ar n+ 7

1

1

a

Operating C3 - C2 and C2 - C1 and

2

= a(a - 1) - 1(a - 1) + 1(1 - a)

using log m - log n = log

= a (a - 1)(a + 1) - 1(a - 1) - 1(a - 1) For infinite solutions, D = 0 2

Þ (a - 1)[a + a - 1 - 1] = 0 2

Þ (a - 1)[a + a - 2] = 0 Þ (a – 1) [a + 2a - a - 2] = 0 (a - 1) = 0, a + 2 = 0 Þ a = – 2, 1; But a ¹ 1 . \a=–2 (d) Applying, C1 ® C1 + C2 + C3 we get 2

2

2

2

2

2

2

2

2

2

(1 + c x )

2

2

2

2

1 + c2 x

1 + (a + b + c + 2) x (1 + b ) x (1 + c ) x

f (x) = 1 + (a + b + c + 2) x

1+ b x

1 + (a + b + c + 2) x (1 + b ) x

1 + b2 x

[As given that 2

2

1 + c2 x

a2 +

b2 +

log r log r

10. (d) Given A2 - A + I = 0 A-1 A2 - A-1 A + A-1.I = A-1.0

(Multiplying A-1 on both sides) -1 Þ A - 1 + A-1 = 0 or A = I - A .

1

11.

1

1

1 (d) Given, D = 1 1 + x 1 1 1+ y

Apply R2 ® R2 – R1 and R ® R3 – R1

Hence, D is divisible by both x and y

c2 =

–2]

2

\ a +b +c +2=0 Applying R1 ® R1 - R2 , R2 ® R2 - R3 0

x -1

0

\ f(x) = 0

1- x

x -1

1 (1 + b 2 ) x 1 + c 2 x

f (x) = ( x - 1)2 Hence degree = 2.

log r log r

n+5

\ D = 0 x 0 = xy 0 0 y

(1 + c 2 x)

1 (1 + b2 ) x

= log ar n+ 2

1 1 1

1 (1 + b2 ) x (1 + c 2 ) x

= 1

log r log r

= 0 (two columns being identical)

Þ (a - 1)[a (a + 2) - 1(a + 2)] = 0

8.

log ar n -1 log ar

2

m we get n

é5 5a a ù a 5a ú and | A2 | = 25 êë0 0 5 úû

12. (a) Given A = ê0

é5 5a a ù é5 5a a ù \ A2 = ê0 a 5a ú ê0 a 5a ú êë0 0 5 úû êë0 0 5 úû

é 25 25a + 5a 2 ê =ê0 a2 0 ê0 ë

5a + 25a 2 + 5a ù ú 5a 2 + 25a ú 25 ú û

EBD_7764 M-106

Mathematics

\ | A2 | = 25 (25a 2 ) \ 25 = 25 (25a 2 ) Þ | a | =

13.

14.

é1 0 ù A= ê ú ë0 –1û Clearly A ¹ I and A ¹ –I and det A = –1 \ Statement 1 is true. Also if A ¹ I then tr(A) = 0 \ Statement 2 is false. (d) The given equations are –x + cy + bz = 0 cx –y + az = 0 bx + ay – z = 0 Q x, y, z are not all zero \ The above system should not have unique (zero) solution

ÞD=0Þ c b

15.

16.

1 5

éa b ù 2 (d) Let A = ê ú then A = I c d ë û Þ a2 + bc = 1 and ab + bd =0 ac + cd = 0 and bc + d2 = 1 From these four relations, a2 + bc = bc + d2 Þ a2 = d2 and b(a + d) = 0 = c( a + d) Þ a = – d We can take a = 1, b = 0, c = 0, d = –1 as one possible set of values, then

–1

a + 1 a -1

a

c

b

–1

a =0

a

–1

Þ –1(1– a2) – c(– c – ab) + b(ac + b) = 0 Þ –1 + a2 + b2 + c2 + 2abc = 0 Þ a2 + b2 + c2 + 2abc = 1 (c) Q All entries of square matrix A are integers, therefore all cofactors should also be integers. If det A = ± 1 then A–1 exists. Also all entries of A–1 are integers. (a) We know that | adj (adj A) | = | A | n–2 A. = | A |0 A = A n–1 Also | adj A | = | A | = | A | 2–1 = | A | \ Both the statements are true but statement-2 is not a correct explanation for statement-1 .

17. (b)

-b b + 1 b - 1 + c -1 c +1

c

a +1

b +1

c -1

a -1

b -1

c +1 = 0

(-1)

a Þ

n+2

a (-1)

n +1

b (-1)n c

a + 1 a -1

-b b + 1 b - 1 + c

c -1 c +1

a +1 a -1

( -1)n + 2 a

b +1 b -1

( -1)n +1 b

c -1 c + 1

( -1) n c

=0

(Taking transpose of second determinant) C1 Û C3 a Þ

a + 1 a -1

-b b + 1 b - 1 c

c -1 c +1

( -1)n + 2 a

a -1 a + 1

( -1)n+ 2 (-b)

b -1 b + 1 = 0

(-1)n + 2 c

c + 1 c -1

C2 Û C3 a

a + 1 a -1

-b b + 1 b - 1 + (-1)n+ 2

Þ

c a

c -1 c +1

a + 1 a -1

-b b + 1 b - 1 = 0 c Þ

c -1 c + 1

a a + 1 a -1 é1 + (-1)n+ 2 ù -b b + 1 b - 1 = 0 ë û c c -1 c + 1

C2 – C1, C3 – C1

M-107

Determinants

a Þ éë1 + (-1)

n+ 2 ù

û

1

-1

-b 2b + 1 2b - 1 = 0 R1 + R3

kx + 3 y - kz = 0

c

3x + y - z = 0

-1

1

Þ

a+c 0 0 é1 + ( -1) n+ 2 ù -b 2b + 1 2b - 1 = 0 ë û -1 c 1

Þ Þ Þ Þ

[1+ (– 1)n + 2](a + c) (2b + 1+ 2b – 1) = 0 4b (a + c) [1 + (–1)n + 2] = 0 1 + (–1)n + 2 = 0 as b (a + c) ¹ 0 n should be an odd integer.

æa b ö 18. (b) Let A = ç ÷ where a, b, c, d ¹ 0 èc d ø æa A2 = ç èc

b öæa ÷ç d øèc

bö ÷ dø

æ a 2 + bc ab + bd ö ÷ Þ A2 = ç ç ac + cd bc + d 2 ÷ è ø

The given system of equations will have non trivial solution, if 1 -k 1 k 3 -k =0 3 1 -1 Þ 1(-3 + k ) + k ( - k + 3k ) + 1( k - 9) = 0 Þ Þ Þ

k - 3 + 2k 2 + k - 9 = 0 k2 + k - 6 = 0 k = -3, k = 2 So the equation will have only trivial solution, when k Î R – {2, – 3} 22. (d) Statement-1 : Determinant of skew symmetric matrix of odd order is zero.

( )

T Statement-2 : det A = det (A).

Þ a 2 + bc = 1, bc + d 2 = 1

det (– A) = – (– 1)n det (A).

ab + bd = ac + cd = 0

c ¹ 0 and b ¹ 0 Þ a + d = 0

19. (c)

x - ky + z = 0

21. (a)

23. (b)

where A is a n ´ n order matrix. For reflexive

| A |= ad - bc = -a 2 - bc = -1

( A, A) Î R

1 2 1 D= 2 3 1 =0 3 5 2

A = P -1 AP is true, For P = I, which is an invertible matrix. \ R is reflexive. For symmetry

3 2 1 D1 = 3 3 1 ¹ 0 1 5 2

Þ Given system, does not have any solution. Þ No solution 20. (a) D = 0 4 k 2 Þ k 4 1 =0 2 2 1 Þ 4(4 - 2) - k (k - 2) + 2(2k - 8) = 0 Þ 8 - k 2 + 2k + 4k - 16 = 0

k 2 - 6k + 8 = 0 Þ (k - 4)(k - 2) = 0 Þ k = 4, 2

As ( A, B) ÎR for matrix P

A = P-1 BP Þ PAP-1 = B Þ

B = PAP -1

Þ

B = P –1

( )

-1

A (P–1)

\ (B, A) Î R for matrix P -1 \ R is symmetric. For transitivity

A = P-1 BP and B = P–1CP Þ

(

)

A = P –1 P -1CP P

EBD_7764 M-108

Þ Þ

( ) CP A = (P ) C (P ) A= P

-1 2

2 -1

2

2

2 \ (A, C) ÎR for matrix P \ R is transitive. So R is equivalence

26.

Mathematics If |P2 + Q2| ¹ 0 then P2 + Q2 is invertible. Þ P–Q=0 Þ P=Q Which gives a contradiction (Q P ¹ Q) Hence |P2 + Q2| = 0 (b) From the given system, we have

k +1 8 4k = ¹ k k + 3 3k - 1 (Q System has no solution) Þ k2 + 4k + 3 = 8k Þ k = 1, 3

æ1ö æ0ö ç ÷ ç ÷ 24. (d) Let Au1 = ç 0 ÷ and Au2 = ç 1 ÷ ç0÷ ç0÷ è ø è ø

If k = 1 then

æ 1ö æ0ö ç ÷ ç ÷ Then, Au1 + Au2 = ç 0 ÷ + ç 1 ÷ ç0÷ ç0÷ è ø è ø

Þ

æ 1ö ç ÷ A ( u1 + u2 ) = ç 1 ÷ ç0÷ è ø

and if k = 3 then ...(1)

æ1 0 0ö ç ÷ Also, A = ç 2 1 0 ÷ ç 3 2 1÷ è ø Þ |A| = 1(1) – 0 (2) + 0 (4–3) = 1 We know,

A-1 =

(Q

A = 1)

Now, from equation (1), we have u1 + u2 = A

æ 1ö ÷ ç 1÷ ç0÷ è ø

=

-1 ç

é 1 0 0ù æ 1 ö é 1 ù ç ÷ = êê -2 1 0 úú ç 1 ÷ = êê -1úú êë 1 -2 1úû èç 0 ø÷ êë -1úû 25. (c) Given P3 = Q3 and P2Q = Q2P Subtracting (1) and (2), we get P3 – P2Q = Q3 – Q2P Þ P2 (P–Q) + Q2 (P – Q) = 0 Þ (P2 + Q2) (P–Q) = 0

...(1) ...(2)

8 4.3 ¹ which is true, 6 9 -1

therefore k = 3 Hence for only one value of k. System has no solution. 27. (b) | P | = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6 Now, adj A = P Þ | adj A | = | P | Þ | A |2 = | P | Þ | P | = 16 Þ 2a – 6 = 16 Þ a = 11 28. (a) Consider

1 adjA A

Þ A -1 = adj ( A )

8 4.1 ¹ which is false 1+ 3 2

3 1 + f (1)

1 + f (1) 1 + f (2)

1 + f (2) 1 + f (3)

1 + f (2)

1 + f (3)

1 + f (4)

1+1 +1

1+ a + b

1 + a 2 + b2

1+ a + b

1 + a 2 + b2

1 + a 3 + b3

1 + a 2 + b2

1 + a3 + b3

1 + a 4 + b4

1 = 1

1 a

1 b

1 ´ 1

1 a

1 b

1

a2

b2

1

a2

b2

1

1

1

1

a

b

1

a2

b2

=

2

= [(1 – a) (1 – b) (a – b)]2 So, K = 1

M-109

Determinants

29. (d) BB' = B ( A-1 A ') ' = B ( A ') '( A-1 ) ' = BA (A–1)' = ( A -1 A ')( A( A -1) ') = A–1A .A'.(A–1)' {as AA' = A'A} = I(A–1A)' = I.I = I 2 = I 30. (a)

æ b - 2 öæ 8 ö ç ÷ç ÷ = –1 è a + 2 øè 3 ø Þ 3a + 8b = 1 Solving (1) and (2), we get \

(2 – l)x1 – 2x2 + x3 = 0 2x1 – (3 + l) x2 + 2x3 = 0 – x1 + 2x2 – lx3 = 0 For non-trivial solution, D=0 2

-(3 + l )

i.e.

Þ

a = 2, b =

1

33. (a)

1 l -1 l - 1 -1 = 0

32. (a)

1

-l

Þ -ll+ ( 1)(l- 1) = 0 Þl= 0, +1, –1 We have k 1 5 2 -k

34. (b)

1 = 28 1

-13 ± 1089 \ 10 since k is an integer, \ k = 2 Also 5k2 + 13k + 66 = 0 k=

Þ k=

-13 ± -1151 10

1 1 1 1 a 1 D= =0 a b 1

Þ 1 [a – b] – 1 [1 – a] + 1 [b – a2] = 0 Þ (a – 1)2 = 0 Þa=1 For a = 1, First two equations are identical ie. x + y + z = 1 To have no solution with x + by + z = 0 b= 1 So b = {1} Þ It is singleton set. Given 2w + 1 = z; z = 3i Þ w=

Þ 5k2 + 13k – 46 = 0 or 5k2 + 13k + 66 = 0 Now, 5k2 + 13k – 46 = 0 Þ

1 2

3i - 1 2 Þ w is complex cube root of unity Applying R1 ® R1 + R2 + R3

-3k 1 k 2

... (2)

æ 1ö orthocentre is çè 21 ÷ø 2

2 =0

-1 2 -l Þ (2 – l) [l(3 + l) – 4] + 2[–2l + 2] + 1[4 – (3 + l)] = 0 Þ l3 + l2 – 5l + 3 = 0 Þ l = 1, 1, 3 Hence l has 2 values. 31. (b) For trivial solution,

1

æ b - 2 öæ 8 ö ç ÷ç ÷ = –1 è a - 5 øè -4 ø a – 2b = 1

Also CH ^ AB

Þ

-2

\

...(1)

2x1 - 2x 2 + x 3 = lx1ü ï 2x1 - 3x 2 + 2x 3 = lx 2 ý - x1 + 2x 2 = lx3 ïþ

2-l

So no real solution exist For orthocentre BH ^ AC

3

k=

-23 ;k = 2 5

=

0

0

2

1 -w - 1 w2 1

w2

w

= 3 (–1 – w – w) = –3 (1 + 2w) = – 3z Þk=–z

EBD_7764 M-110

Mathematics

10-13 delet

Continuity and Differentiability 1.

2.

f is defined in [-5, 5] as [2002] f (x) = x if x is rational = – x if x is irrational. Then (a) f (x) is continuous at every x, except x = 0 (b) f (x) is discontinuous at every x, except x = 0 (c) f (x) is continuous everywhere (d) f (x) is discontinuous everywhere

3.

If y = (x + 1 + x

[2002] (b) 1 (d) 2

2 )n, then (1 + x2)

(a) n2y (c) –y 4.

(d) neither differentiable nor continuous at x = 0 6.

d2y

(a) 1 7.

+x

8.

5.

[2003]

Let f ( x) =

(c)

(a) discontinuous every where [2003] (b) continuous as well as differentiable for all x (c) continuous for all x but not differentiable at x=0

1 - tan x p é pù , x ¹ , x Î ê 0, ú . 4x - p 4 ë 2û

(a) –1

(b) 4 (d) 1

then f(x) is

[2003]

é pù æ pö If f (x) is continuous in ê 0, ú , then f çè ÷ø is 4 ë 2û [2004]

f ( a ) g ( x) - f ( a ) - g ( a ) f ( x ) + f ( a ) =4 g ( x) - f ( x)

ì æ 1 1ö ï -ç + ÷ If f ( x ) = í xe è x x ø , x ¹ 0 ï0 ,x = 0 î

(d) 0. (c) 2 n - 1 Let f (x) be a polynomial function of second

(a) Arithmetic -Geometric Progression (b) A.P (c) G..P (d) H.P.

dx2 (b) – n2y [2002] (d) 2x2y

n

then the value of k is (a) 0 (c) 2

(b) 2 n

f '(a), f ¢ (b), f '(c) are in

dy is dx

n. Further if

lim

f ' (1) f ' ' (1) f ' ' ' (1) ( -1) n f n (1) + + .......... n! 1! 2! 3!

degree. If f(1) = f(-1) and a, b, c are in A. P , then

f (a) , g (a) exist and are not equal for some

x ®a

[2003]

is

Let f (a) = g (a) = k and their nth derivatives n

If f ( x) = x n , then the value of f (1) -

If f ( x + y ) = f ( x ). f ( y )"x. y and f (5) = 2, f '(0) = 3, then f ¢ (5) is (a) 0 (c) 6

20

9.

-

(b) 1 2

If x = e y + e

1 2

(d) 1 y +L to ¥

, x > 0, then

dy is dx

(a)

1+ x x

(b)

1 x

(c)

1- x x

(d)

x 1+ x

[2004]

M-111

Continuity and Differentiability

10.

Suppose f (x) is differentiable at x = 1 and

lim

1

h®0 h

11.

f (1 + h) = 5 , then f '(1) equals [2005]

(a) 3 (b) 4 (c) 5 (d) 6 Let f be differentiable for all x. If f (1) = – 2 and f '( x) ³ 2 for x Î [1, 6], then

12.

[2005]

(a) f (6) ³ 8 (b) f (6) < 8 (c) f (6) < 5 (d) f (6) = 5 If f is a real valued differentiable function satisfying | f (x) – f (y) | £ ( x - y )2 , x, y Î R and

13.

14.

f (0) = 0, then f (1) equals [2005] (a) – 1 (b) 0 (c) 2 (d) 1 The value of a for which the sum of the squares of the roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the least value is [2005] (a) 1 (b) 0 (c) 3 (d) 2 The set of points where f ( x ) = differentiable is (a)

(-¥,0) È (0, ¥)

(b)

(-¥,-1) È (-1, ¥)

(c)

(-¥, ¥)

(d)

(0, ¥)

x is 1+ | x | [2006]

dy is dx

15.

If x m . y n = ( x + y ) m+ n , then

16.

x+ y xy x (c) xy (d) y Let f : R ® R be a function defined by

(a)

y x

[2006]

(b)

(c) f (x) ³ 1 for all x Î R (d) f (x) is not differentiable at x = 1 17. The function f : R /{0} ® R given by [2007]

1 2 - 2x x e -1 can be made continuous at x = 0 by defining f (0) as (a) 0 (b) 1 (c) 2 (d) – 1 f ( x) =

1 ì if x ¹ 1 ï( x –1) sin 18. Let f ( x) = í [2008] x –1 ïî 0 if x = 1 Then which one of the following is true? (a) f is neither differentiable at x = 0 nor at x =1 (b) f is differentiable at x = 0 and at x =1 (c) f is differentiable at x = 0 but not at x = 1 (d) f is differentiable at x = 1 but not at x = 0 19. Let y be an implicit function of x defined by x2x – 2xx cot y – 1= 0. Then y¢(1) equals [2009] (a) 1 (b) log 2 (c) –log 2 (d) –1 20. Let f : (–1, 1) ® R be a differentiable function with f (0) = – 1 and f ¢ (0) = 1. Let g(x) = [f (2f (x) + 2)]2. Then g¢(0) = [2010] (a) –4 (b) 0 (c) –2 (d) 4 21. The values of p and q for which the function [2011] ì ï sin( p + 1) x + sin x , x < 0 x ï ï f ( x) = í q , x = 0 is continuous for ï 2 ï x+ x - x ,x > 0 ïî x3 / 2

all x in R, are (a)

5 1 p = ,q = 2 2

(b)

3 1 p = - ,q = 2 2

(c)

1 3 p = ,q = 2 2

(d)

1 3 p = ,q = 2 2

f (x) = min {x + 1, x + 1} ,Then which of the following is true ? (a) f (x) is differentiable everywhere [2007] (b) f (x) is not differentiable at x = 0

EBD_7764 M-112

22.

d2x dy 2

Mathematics

equals :

[2011]

(b)

a f (a ) - a 2 f ' ( a )

(c)

2af ( a) - a 2 f ' ( a)

(d)

2a f (a) +a2 f '(a)

-1

(a)

æ d 2 y ö æ dy ö-3 -ç 2 ÷ ç ÷ ç dx ÷ è dx ø è ø

25. If f : R ® R is a function defined by f (x) = [x]

æ d 2 y ö æ dy ö -2 (b) çç 2 ÷÷ ç dx ÷ è dx ø è ø (c)

æ d 2 y ö æ dy ö-3 -ç 2 ÷ç ÷ ç dx ÷ è dx ø è ø

æ d2y ö (d) çç 2 ÷÷ è dx ø

23.

-1

Define f (x) as the product of two real function [2011RS] ì 1 ïsin , if x ¹ 0 f1 ( x ) = x, x Î R, and f 2 ( x ) = í x ï if x = 0 î0,

as follows : ïì f1 ( x ) . f 2 ( x) , if x = 0 f ( x) = í if x = 0 ïî 0 Statement - 1 : f (x) is continuous on R. Statement - 2 : f1 ( x) and f 2 ( x) are continuous

24.

on R. (a) Statement -1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true If function f (x) is differentiable at x = a, x f (a) - a f ( x) then xlim is : ®a x-a 2

(a)

-a 2 f ' ( a )

2

[2011RS]

æ 2x - 1 ö cos ç ÷ p , where [x] denotes the greatest è 2 ø integer function, then f is . [2012] (a) continuous for every real x. (b) discontinuous only at x = 0 (c) discontinuous only at non-zero integral values of x. (d) continuous only at x = 0. 26. Consider the function, f (x) = | x – 2 | `+ | x – 5 |, x Î R. Statement-1 : f ¢(4) = 0 Statement-2 : f is continuous in [2,5], differentiable in (2,5) and f (2) = f (5). [2012] (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false.

27. If y = sec(tan–1x), then

dy at x = 1 is equal to : dx

[2013] (a)

1 2

(c) 1

(b)

1 2

(d)

2 28. If f and g are differentiable functions in [0, 1] satisfying f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c Î]0,1[ (a) f ¢(c) = g¢(c) (c) 2f ¢(c) = g¢(c)

[2014] (b) f ¢(c) = 2g ¢(c) (d) 2f ¢(c) = 3g¢(c)

M-113

Continuity and Differentiability

29.

If the function. ïì k x + 1, 0 £ x £ 3 g(x) = í is differentiable, ïî m x + 2, 3 < x £ 5

then the value of k + m is : (a)

10 3

æ pö then y ç ÷ is equal to : è 2ø

(b) 4

(c) 2 30.

[2015]

(b) g is differentiable at x = 0 and g'(0) = – sin(log2) (c) g is not differentiable at x = 0 (d) g'(0) = cos(log2) dy 31. If (2 + sin x) + (y + 1) cos x = 0 and y(0) = 1, dx

16 5

(d)

For x Î R, f(x) = |log2 – sinx| and g(x) = f(f(x)), then : [2016] (a) g'(0) = – cos(log2)

(a)

4 3

(c)

-

(b) 2 3

1 3

(d) -

1 3

Answer Key 1 (b) 16 (a) 31 (b)

1.

2 (c) 17 (b)

3 (a) 18 (c)

4 (b) 19 (d)

5 (c) 20 (a)

6 (d) 21 (b)

7 (b) 22 (c)

8 (c) 23 (c)

(b) Let a is a rational number other than 0, in [–

9 (c) 24 (c)

2.

5, 5], then f (a) = a and lim f ( x) = -a x®a

[As in the immediate neighbourhood of a rational number, we find irrational numbers] \ f (x) is not continuous at any rational number If a is irrational number, then f (a) = – a and lim f ( x ) = a x® a

\ f (x) is not continuous at any irrational number clearly lim f ( x) = f (0) = 0 x®0

\ f (x) is continuous at x = 0

10 (c) 25 (a)

11 (a) 26 (c)

12 (b) 27 (a)

13 (a) 28 (b)

14 (c) 29 (c)

15 (a) 30 (d)

(c) f (x + y) = f (x) ´ f (y) Differentiate with respect to x, treating y as constant f ¢ (x + y) = f ¢ (x) f (y) Putting x = 0 and y = x, we get f '(x)= f '(0) f (x) ;

3.

(a)

Þ f ¢ (5) = 3 f (5) = 3 × 2 = 6.

y = ( x + 1 + x 2 )n dy æ 1 ö = n( x + 1 + x 2 )n -1 ç 1 + (1 + x 2 ) -1/ 2 . 2 x; ÷ è 2 ø dx

dy ( 1 + x 2 + x) = n( x + 1 + x 2 ) n -1 dx 1 + x2

EBD_7764 M-114

Mathematics

=

n( 1 + x 2 + x ) n

7.

1 + x2 2 dy = ny or or 1 + x dx dy ( y1 = ) dx

(b)

f (1) = f (-1) Þ a + b + c = a - b + c or b = 0

1 + x 2 y 1= ny

\ f ( x) = ax 2 + c or f '( x ) = 2ax

Squaring,

Now f '(a); f '(b); and f '( c) are 2a(a); 2a(b); 2a(c ) i.e. 2a2, 2ab, 2ac. Þ If a, b, c are in A.P. then f '(a); f '(b) and f '(c ) are also in A.P..

(1 + x 2 ) y12 = n2 y 2 Differentiating, (1 + x 2 )2 y1 y2 + y12 .2 x = n2 .2 yy 1 or (1+x2)y2 + xy1 = n2y

4.

5.

(b)

(c)

f (a) g ¢ ( x ) - g (a ) f ¢ ( x ) =4 g ¢( x ) - f ¢ ( x ) x® a (By L’ Hospital rule) k g ¢ ( x) - k f ¢ ( x) lim =4 x ® a g ¢ ( x) - f ¢ ( x) \ k = 4. lim

f (0) = 0; f ( x ) = xe

æ 1 1ö -ç + ÷ è x xø

h

h ®0 e 2 / h

h® 0

æ 1 1ö -ç - ÷ è h hø

h® 0

=0

therefore, f (x) is continuous. R.H.D = lim

(0 + h)e

L.H.D. = lim

-0

h

h® 0

(0 - h)e

æ 1 1ö -ç - ÷ è h hø

-0

=1

Q f (x) is continuous at x =

n + 1!

4

2!

( x ) = n ! Þ f (1) = n! n ( n - 1)( n - 2 ) n n! + ¼ + ( -1) n

3!

=n C0 -n C1 + n C2 - n C3 + ¼ + ( -1)

n!

n n

Cn = 0

p 4

1 - tan x æ pö \ f çè ÷ø = limp f ( x ) = limp 4x - p 4 x® x®

.........................

= 1-

1 + tan h 1 - tan h 4h

tan h -2 1 -2 . = =4 2 h ®0 1 - tanh 4 h

f '' ( x ) = n ( n - 1) x n- 2 Þ f '' (1) = n ( n - 1)

n ( n - 1)

4

= lim

f ' ( x ) = nx n-1 Þ f ' (1) = n

......................... f

x®

4

h® 0

f ( x ) = x n Þ f (1) = 1

n

1 - tan x is continuous in 4x - p

é pù ê0, 2 ú ë û æ pö \ f ç ÷ = lim p f ( x) = lim + f ( x ) è 4 ø x® p

= lim

=0

-h therefore, L.H.D. ¹ R.H.D. f (x) is not differentiable at x = 0.

(d)

f ( x) =

1-

h® 0

6.

(c)

æp ö 1 - tan ç + h÷ è4 ø = lim ,h > 0 ö h ®0 æ p 4 ç + h÷ - p è4 ø

=0

æ 1 1ö -ç + ÷ è h hø

8.

æp ö = lim f ç + h÷ ø h ®0 è 4

R.H.L. lim (0 + h)e -2 / h = lim L.H.L. lim (0 - h)e

f ( x) = ax 2 + bx + c

4

- sec2 x [using L' Hospital’s rule] 4 x® 4 p - sec 2 4 = -2 = -1 = 4 4 2

= limp

M-115

Continuity and Differentiability

9.

(c)

y +L¥

x = e y+ e Taking log.

Þ x = e y+ x .

14. (c)

1 dy = +1 x dx dy 1 1- x \ = -1 = dx x x

log x = y + x Þ

10.

(c)

Hence f '(1) = lim

h®0

11.

12.

f (1 + h) =5 h

(a) As f (1) = – 2 & f '( x) ³ 2 " x Î [1, 6] Applying Lagrange’s mean value theorem f (6) - f (1) = f '(c) ³ 2 5 Þ f (6) ³ 10 + f (1) Þ f (6) ³ 10 – 2 Þ f (6) ³ 8. (b)

f '(x) = lim

h®0

| f '( x) | = lim

f ( x + h) - f ( x ) h

h®0

£ lim

h®0

f ( x + h ) - f ( x) h

(a)

\ 15. (a)

= a2 - 2a + 6 = (a - 1)2 + 5 For min. value of a2 + b2 where a is an integer Þ a = 1.

x³0

ì x , ï 2 ï (1 - x ) f '( x) = í ï x , ïî (1 + x )2

x 0 and q > 0. Then which one of the following holds? [2008] (a) The cubic has minima at p 3

at –

(b) The cubic has minima at – at

p and maxima 3

p and maxima 3

p 3

(c) The cubic has minima at both –

p 3

(d) The cubic has maxima at both

y = x+

p and 3

p 3 How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have? [2008] (a) 7 (b) 1 (c) 3 (d) 5 –

22.

p and 3

M-121 23. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is [2009] 2 2 (a) (x – 2)y' = 25 –(y – 2) (b) (y – 2)y'2 = 25 –(y – 2)2 (c) (y – 2)2y'2 = 25 –(y – 2)2 (d) (x – 2)2 y'2 = 25 –(y – 2)2 24. Let f (x) = x | x | and g (x) = sin x. Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0. [2009] (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. 25. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P' (x) = 0. If P(–1) < P(1), then in the interval [ –1, 1] : [2009] (a) P(–1) is not minimum but P(1) is the maximum of P (b) P(–1) is the minimum but P(1) is not the maximum of P (c) Neither P(–1) is the minimum nor P(1) is the maximum of P (d) P(–1) is the minimum and P(1) is the maximum of P 26. The equation of the tangent to the curve

4

, that is parallel to the x-axis, is [2010] x2 (a) y = 1 (b) y = 2 (c) y = 3 (d) y = 0 27. Let f : R ® R be defined by f ( x) =

{ k2-x+23,x, ifif xx £>--11

If f has a local minimum at x = – 1 , then a possible value of k is [2010] -

(a) 0

(b)

(c) –1

(d) 1

1 2

EBD_7764 M-122 28. Let f : R ® R be a continuous function defined

1 by f ( x ) = x e + 2e - x Statement -1 : f (c) =

[2010] 1 , for some c Î R. 3

Statement -2 : 0 < f (x) £

29.

1 2 2

, for all x Î R

(a) Statement -1 is true, Statement -2 is true ; Statement -2 is not a correct explanation for Statement -1. (b) Statement -1 is true, Statement -2 is false. (c) Statement -1 is false, Statement -2 is true . (d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1. Let f be a function defined by [2011RS] ì tan x , x¹0 ï f ( x) = í x ï1, x=0 î

Statement - 1 : x = 0 is point of minima of f Statement - 2 : f ¢ ( 0 ) = 0.

30.

(a) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. The curve that passes through the point (2, 3), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact is given by : [2011RS] (a) 2 y - 3x = 0

(b) y = 2

(c) x2 + y 2 = 13

6 x 2

æxö æ yö (d) ç ÷ + ç ÷ = 2 è 2ø è 3ø

Mathematics 31. A spherical balloon is filled with 4500p cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72p cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is:

(a)

9 7

(b)

7 9

[2012]

2 9 (d) 9 2 32. Let a, b Î R be such that the function f given by f (x) = ln | x | + bx2 + ax, x ¹ 0 has extreme values at x = –1 and x = 2 Statement-1 : f has local maximum at x = –1 and at x = 2.

(c)

1 -1 and b = [2012] 2 4 (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false. 33. The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] [2013] (a) lies between 1 and 2 (b) lies between 2 and 3 (c) lies between .1 and 0 (d) does not exist. 34. If x = –1 and x = 2 are extreme points of

Statement-2 : a =

f ( x ) = a log x + b x 2 + x then

(a)

a = 2, b = -

1 2

(c)

a = -6, b =

1 2

(b)

[2014] a = 2, b =

1 2

(d) a = -6, b = -

1 2

Application of Derivatives

35.

36.

37.

The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) [2015] (a) meets the curve again in the third quadrant. (b) meets the curve again in the fourth quadrant. (c) does not meet the curve again. (d) meets the curve again in the second quadrant. Consider æ 1 + sin x ö æ ö ÷, x Î ç0,p÷. f (x) = tan -1 ç ç ÷ ç 1 - sin x ÷ è 2ø è ø p A normal to y = f(x) at x = a so passes through 6 the point : [2016] æp ö æp ö (a) ç (b) ç ç , 0÷ ÷ ç , 0÷ ÷ è6 ø è4 ø æ 2pö (c) (0, 0) (d) ç ç0, ÷ ÷ è 3 ø A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then: [2016] (a) x = 2r (b) 2x = r

M-123 (c) 2x = (p + 4)r (d) (4 – p) x =pr 38. The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point:

(a)

æ 1 1ö çè , ÷ø 2 3

æ 1 1ö (b) ç - , - ÷ è 2 2ø

æ 1 1ö æ 1 1ö (d) ç , - ÷ çè , ÷ø è 2 3ø 2 2 39. Twenty metres of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is : [2017] (a) 30 (b) 12.5 (c) 10 (d) 25 40. The eccentricity of an ellipse whose centre is 1 at the origin is . If one of its directices is x = – 2 (c)

æ 3ö 4, then the equation of the normal to it at ç1, ÷ è 2ø is : [2017] (a) x + 2y = 4 (b) 2y – x = 2 (c) 4x – 2y = 1 (d) 4x + 2y = 7

Answer Key 1 (b) 16 (a) 31 (c)

1.

2 (a) 17 (c) 32 (b)

3 (d) 18 (c) 33 (d)

4 (c) 19 (d) 34 (a)

5 (a) 20 (c) 35 (b)

6 (b) 21 (a) 36 (d)

(b) Distance of origin from (x, y) = =

æ at ö a 2 + b2 - 2ab cos ç t - ÷ ; è bø

£ a 2 + b 2 + 2ab

éì ù æ at ö ü = -1ú ê í cos ç t - ÷ ý è b ø þ min ëê î ûú

7 (d) 22 (b) 37 (a)

8 (d) 23 (c) 38 (c)

9 (a) 24 (b) 39 (d)

10 (d) 25 (a) 40 (c)

11 (b) 26 (c)

12 (b) 27 (c)

13 (c) 28 (d)

14 (c) 29 (b)

15 (d) 30 (b)

=a+b \ Maximum distance from origin = a + b

x2 + y 2 2.

ax3 bx 2 + + cx 3 2 Þ f (0) = 0 and

(a) Let f (x) =

a b 2a + 3b + 6c + +c = =0 3 2 6 Also f (x) is continuous and differentiable

f (1) =

EBD_7764 M-124

Mathematics

in [0, 1] and [0, 1[. So by Rolle’s theorem, f ¢ (x) = 0. i.e ax2 + bx + c = 0 has at least one root in [0, 1]. 3.

Þ 1 = (2 - 1)3 + D Þ D = 0 \ f (x) = ( x – 1)3

7.

(d) f ( x) = 2 x 3 - 9ax 2 + 12a 2 x + 1

Þ x = a or x = 2a.At x = a max .

dy = - cot q. dx \ The slope of the normal at q = tan q \ The equation of the normal at q is y - a sin q = tan q( x - a - a cos q)

and at x = 2a min \ p = a and q = 2a

Þ y cos q - a sin q cos q = x sin q - a sin q - a sin q cos q

For max. or min. 6 x 2 - 18ax + 12 a 2 = 0 Þ x 2 - 3ax + 2a 2 = 0

As per question p 2 = q 2

5.

6.

\ a = 2a Þ a = 2or a = 0 but a > 0, therefore, a = 2. 1 1 dy (c) y = x + or = 1x dx x2 1 For max. or min., 1 = 0 Þ x = ±1 x2 æ d 2y ö 2 d2y ÷ =2 = Þç 2 3 ç dx 2 ÷ dx x ø x =2 è (+ve minima) \x = 1

dy dy 9 = 18 Þ = dx dx y dy 9 9 Given = 2Þ =2Þ y = dx y 2 9 Putting in y 2 = 18 x Þ x = 8 æ9 9ö \ Required point is ç , ÷ è8 2ø (b) f ¢¢ ( x) = 6( x - 1). Inegrating, we get (a)

dx dy = - a sin q and = a cos q dq dq \

f '( x ) = 6 x 2 - 18ax + 12a 2 ; f ''( x ) = 12 x - 18a

4.

(d)

8.

ax3 bx 2 + + cx 3 2 Being polynomial, it is continuous and differentiable, also, a b f (0) = 0 and f (1) = + + c 3 2 2a + 3b + 6c Þ f (1) = = 0 (given) 6 \ f (0) = f (1) \ f (x) satisfies all conditions of Rolle’s theorem therefore f ¢(x) = 0 has a root in (0, 1) f ( x) =

y 2 = 18 x Þ 2 y

f ¢ ( x) = 3 x 2 - 6 x + c

Slope at (2, 1) = f ¢(2) = c = 3 [Q slope of tangent at (2,1) is 3] 2

\ f ¢ ( x ) = 3x - 6 x + 3 = 3( x - 1) Inegrating again, we get f ( x ) = ( x - 1)3 + D The curve passes through (2, 1)

2

Þ x sin q - y cos q = a sin q Þ y = ( x - a ) tan q which always passes through (a, 0) (d) Let us define a function

9.

i.e. ax 2 + bx + c = 0 has at lease one root in (0, 1) (a) Area of rectangle ABCD = 2a cos q (2b sin q) = 2ab sin 2q Y (-a cos q, b sin q) B

(a cos q, b sin q) A X

C ( - a cos q, - b sin q)

D (a cos q, - b sin q)

Þ Area of greatest rectangle is equal to 2ab When sin 2q = 1 .

M-125

Application of Derivatives

10.

(d)

x = a ( cos q + q sin q)

13. (c) Clearly function f (x) = 3 x 2 - 2 x + 1 is

dx = a ( - sin q + sin q + q cos q ) dq dx = aq cos q .....(1) Þ dq y = a ( sin q - q cos q)

increasing when f '( x ) = 6x – 2 ³ 0

Þ

dy = a [ cos q - cos q + q sin q ] dq dy = aq sin q .....(2) Þ dq From equations (1) and (2) we get dy = tan q Þ Slope of normal = – cot q dx Equation of normal at 'q ' is y – a (sin q – q cos q ) = – cot q (x – a (cos q + q sin q )

11.

Þ y sin q – a sin 2 q + a q cos q sin q = – x cos q + a cos 2 q + a q sin q cos q Þ x cos q + y sin q = a Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin. (b) Given that d æ 4 3ö dv = 50 cm3/min Þ ç pr ÷ø = 50 dt è 3 dt 2 Þ 4 pr

dr = 50 dt 50

dr 1 cm/min Þ dt = 4 p(15) 2 = 18p (here r = 10+5)

12.

(b) Let f (x) = an x n + an -1 x n -1 + ...... + a1 x = 0 The other given equation, n -1

n -2

+ (n – 1) an -1 x nan x f ¢(x) Given a1 ¹ 0 Þ f (0) = 0

+ ....+ a1 = 0 =

Again f (x) has root a, Þ f (a ) = 0 \ f (0) = f (a) \ By Rolle’s theorem f ¢(x) = 0 has root between ( 0, a ) Hence f ¢ ( x ) has a positive root smaller than a.

Þ x Î[1/ 3, ¥) \ f (x) is incorrectly matched with 1ù æ ç -¥, 3 ú è û 14. (c) Let the lizard catches the insect after time t then distance covered by lizard = 21cm + distance covered by insect Þ

1 2 ft = 4 ´ t + 21 2

Þ

1 ´ 2 ´ t 2 = 20 ´ t + 21 2

Þ t 2 - 20t - 21 = 0 Þ t = 21sec

15. (d)

A

B

u=0

f t+m

s+n

v

u=0

s f¢ v t As per question if point B moves s distance in t time then point A moves (s + n) distance in time (t + m) after which both have same velocity v. Then using equation v = u + at we get

v = f (t + m) = f ' t Þ t =

f m f '- f

....(1)

Using equation v 2 = u 2 + 2 , as we get v2 = 2 f ( s + n ) = 2 f ' s

Þs=

f n f '- f ....(2)

Also for point B using the eqn 1 s = ut + at 2 , we get 2 1 f 't 2 2 Substituting values of t and s from equations (1) and (2) in the above relation, we get s=

EBD_7764 M-126

Mathematics 2

2

f n f m 1 = f' f '- f 2 ( f '- f ) 2 Þ ( f '- f ) n =

16.

(a)

1 ff ' m2 2

1 x 2 + is of the form y + where y 2 x 1 y + ³ 2 and equality holds for y = 1 y x \ Min value of function occurs at = 1 2 i.e., at x = 2 x 2 1 2 f (x) = + Þ f '( x ) = - 2 = 0 2 x 2 x

Þ x2 = 4 or x = 2, – 2;

f ''( x) =

4 x3

f ''( x )] x = 2 = + ve Þ f ( x ) has local min at x = 2.

17.

(c) Area =

1 2 x sin q 2 x

q

x

Maximum value of sinq is 1 at q =

18.

p 2

1 2 x 2 (c) Using Lagrange's Mean Value Theorem Let f (x) be a function defined on [a, b] f (b ) - f ( a ) then, f '(c) = ....(i) b-a c Î [a, b] 1 \ Given f (x) = logex \ f ' (x) = x \ equation (i) become 1 f (3) - f (1) = c 3-1 1 loge 3 - loge 1 log e3 = = Þ c 2 2 Amax =

2 loge 3 Þ c = 2 log3e 19. (d) Given f (x) = tan–1 (sin x + cos x) 1 .(cos x - sin x ) f '(x) = 1 + (sin x + cos x) 2 Þ

=

=

\

c=

æ 1 ö 1 2. ç cos x sin x÷ è 2 ø 2 1 + (sin x + cos x )2

p p æ ö çè cos .cos x - sin .sin x÷ø 4 4 1 + (sin x + cos x) 2

f '(x) =

pö æ 2 cos ç x + ÷ è 4ø

1 + (sin x + cos x ) 2 if f ' (x) > 0 then f (x) is increasing function. Hen ce f (x) is increasing, if p p p - < x+ < 2 4 2 3p p Þ 0, " x ÎR Þ f is an increasing function on R x=–

22.

Also lim f ( x) = ¥ and x ®¥

lim f ( x ) = – ¥

x® – ¥

23.

ìïsin (– x 2 ), if x < 0 =í 2 ïîsin ( x ), if x ³ 0

Þ The curve y = f (x) crosses x-axis only once. \ f (x) = 0 has exactly one real root. (c) Let the centre of the circle be ( h, 2) \ Equation of circle is …(1) ( x – h) 2 + ( y – 2) 2 = 25 Differentiating with respect to x, we get dy 2( x – h) + 2( y – 2) = 0 dx dy Þ x – h = –( y – 2) dx Substituting in equation (1) we get

2 ïì – 2 x cos x , if x < 0 \ (go f )¢ (x) = í 2 ïî 2 x cos x , if x ³ 0 Here we observe L (go f )¢ (0) = 0 = R (go f )¢ (0) Þ go f is differentiable at x = 0 and (go f )¢ is continuous at x = 0 Now (go f )'' (x)

2 2 2 ïì – 2 cos x + 4 x sin x , x < 0 = í 2 2 2 ïî 2 cos x - 4 x sin x , x ³ 0 Here L(go f )'' (0) = – 2 and R (go f )'' (0) = 2 Q L(go f)'' (0) ¹ R (go f )'' (0) Þ go f (x) is not twice differentiable at x = 0. \ Statement - 1 is true but statement -2 is false. 25. (a) We have P (x) = x4 + ax3 + bx2 + cx + d Þ P' (x) = 4 x3 + 3ax2 + 2bx + c But P' (0) = 0 Þ c = 0 \ P(x) = x4 + ax3 + bx2 + d As given that P ( – 1) < P (a) Þ 1–a+b +d < 1+a+b+d Þ a>0 Now P ' (x) = 4x3 + 3ax2 +2bx = x (4x2 + 3ax + 2b) As P' (x) = 0, there is only one solution x = 0, therefore 4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0

9a 2 >0 32 Hence a, b > 0 Þ P' (x) = 4 x 3 + 3ax2 + 2bx > 0 "x > 0

Þ 9a2 – 32 b < 0

Þ

b>

EBD_7764 M-128

Mathematics

\ \

26.

27.

P (x) is an increasing function on (0,1) P (0) < P (a) Similarly we can prove P (x) is decreasing on (– 1, 0) \ P (– 1) > P (0) So we can conclude that Max P (x) = P (1) and Min P (x) = P (0) Þ P(–1) is not minimum but P (1) is the maximum of P. (c) Since tangent is parallel to x-axis, 8 dy = 0 Þ 1- 3 = 0 Þ x = 2 Þ y = 3 \ dx x Equation of tangent is y – 3 = 0 (x – 2) Þy=3 (c) ì k - 2 x, if x £ -1 f ( x) = í î 2 x + 3, if x > -1 2x + 3

k-2x 1

Y y = tan x y=x

X´

O

X

Y´

In left neighbourhood of ‘0’ tan x < x tan x > 1 as ( tan x < 0) x at x = 0, f ( x) = 1 Þ x = 0 is the point of minima So Statement 1 is true. Statement 2 obvious. dy 30. (b) Y – y = ( X - x) dx Y

–1

B (0, y-xdy/dx)

This is true where k = – 1 28.

(d)

1

f ( x) =

e x + 2e - x

f '( x ) =

=

ex e2 x + 2

(e 2 x + 2) e x - 2e 2 x .e x (e 2 x + 2) 2

f '( x) = 0 Þ e 2 x + 2 = 2e 2 x e2 x = 2 Þ e x = 2 maximum f (x) = 0 < f ( x) £

Since 0 < f ( c) =

29.

(b)

(x, y)

1 2 2

2 1 = 4 2 2

"x Î R

1 1 < Þ for some c Î R 3 2 2

1 3

ì tan x , x¹0 ï f ( x) = í x ï 1, x = 0 î In right neighbourhood of ‘0’ tan x > x tan x >1 x

X´

O

A (x – y)/(dy/ dx), 0) Y´

y dy / dx xdy Y-intercept = y – dx According to given statement y xdy x - = 2 x and y = 2y dy dx - xdy -y =y =x and dy dx dx dx dy Þ + =0 x y lny = -lnc + lnc c y= x

X-intercept = x -

X

M-129

Application of Derivatives

31.

32.

Since the above line passes through the point (2, 3). \ c=6 6 Hence y = is the required equation. x (c) Volume of spherical balloon 4 = V = pr 3 3 4pr 3 Þ 4500 p = 3 (Q Given, volume = 4500pm3) Differentiate both the side, w.r.t 't' we get, dV æ dr ö = 4pr 2 ç ÷ dt è dt ø dV = 72 p Now, it is given that dt \ After 49 min, Volume = (4500 – 49 ´ 72)p = (4500 – 3528)p = 972 p m3 Þ V = 972 p m3 4 \ 972 p = pr 3 3 Þ r3 = 3 ´ 243 = 3 ´ 35 = 36 = (32)3 Þ r=9 dV = 72 p Also, we have dt æ dr ö \ 72 p = 4 p´ 9 ´ 9 ç ÷ è dt ø dr æ 2 ö =ç ÷ Þ dt è 9 ø (b) Given, f ( x ) = ln x + bx 2 + ax 1 \ f ' ( x ) = + 2bx + a x At x = –1, f '(x) = –1– 2b + a =0 Þ a – 2b = 1 ...(i) 1 At x = 2, f '(x) = + 4b + a = 0 2 1 Þ a + 4b = ...(ii) 2 1 1 On solving (i) and (ii) we get a = , b = 2 4 1 x 1 2 - x2 + x Thus, f ' ( x ) = - + = x 2 2 2x 2 2 x x 2 -x + x + 2 = = 2x 2x - ( x + 1)( x - 2 ) = 2x

(

)

+ –¥

+ –1

–

0

2 –

¥

So maxima at x = –1, 2 33. (d) f (x) = 2x3 + 3x + k f'(x) = 6x2 + 3 > 0 " x Î R (Q x2 > 0) Þ f(x) is strictly increasing function Þ f(x) = 0 has only one real root, so two roots are not possible. 34. (a) Let f (x) = a log | x | + bx2 + x Differentiate both side, a f ¢ ( x ) = + 2b x + 1 x Since x = –1 and x = 2 are extreme points therefore f ¢ ( x ) = 0 at these points. Put x = –1 and x = 2 in f ¢( x ) , we get – a –2b + 1 = 0 Þ a +2b = 1 ...(i) a + 4b + 1 = 0 Þ a +8b = –2 ...(ii) 2 On solving (i) and (ii), we get 1 6b = -3 Þ b = 2 \ a=2 35.

(b)

Given curve is x2 + 2xy – 3y2 = 0 Differentiatew.r.t. x

, 2x + 2x

...(1)

dy dy + 2y - 6y =0 dx dx

æ dy ö =1 ç ÷ è dx ø(1, 1) Equation of normal at (1, 1) is y=2–x ...(2) Solving eq. (1) and (2), we get x=1,3 Point of intersection (1, 1), (3, –1) Normal cuts the curve again in 4th quadrant. 36.

æ 1 + sin x ö f ( x) = tan –1 ç è 1– sin x ÷ø

(d)

æ ç = tan –1 ç ç ç è

2ö x xö æ xö æ çè sin + cos ÷ø ÷ 1 + tan ç 2 2 ÷ –1 2÷ 2 ÷ = tan ç x÷ x xö æ ç 1 – tan ÷ çè sin – cos ÷ø ÷ è 2ø 2 ø x

æ æ p xö ö = tan –1 ç tan ç + ÷ ÷ è 4 2ø ø è

EBD_7764 M-130

Mathematics

p x + 4 2 dy 1 Þ = dx 2 Þ y=

Slope of normal =

–1 = –2 æ dy ö çè ÷ø dx

æp p pö , + ÷ 6 4 12 ø pö æp pö æ y – ç + ÷ = –2 ç x – ÷ è 4 12 ø è 6ø 4p 2p y– = –2 x + 12 6 p p y – = –2 x + 3 3 2p y = –2 x + 3

At çè

æ1 1ö \ ç , ÷ satisfy it. è2 2ø 39. (d) We have Total length = r + r + rq = 20 Þ 2r + r q = 20 20 - 2r Þq= ...(1) r 1 1 æ 20 - 2r ö q ´ pr 2 = r 2q = r 2 ç A = Area = ÷ 2p 2 2 è r ø 2 A = 10r – r For A to be maximum q r r dA = 0 Þ 10 – 2r = 0 dr Þr =5 d 2A

This equation is satisfied only by the point

æ 2p ö çè 0, ÷ø 3 37.

(a)

Þ 2x + pr = 1

4x + 2pr = 2 S = x2 + pr2 2

æ 1 - pr ö 2 S=ç ÷ + pr è 2 ø dS æ 1 - pr öæ -p ö = 2ç ÷ç ÷ + 2pr dr è 2 øè 2 ø -p p2 r 1 + + 2pr = 0 Þ r = 2 2 p+ 4 2 Þx= Þ x = 2r p+ 4 x+6 We have y = (x - 2)(x - 3) At y-axis, x = 0 Þ y = 1 On differentiating, we get Þ

38.

(c)

dy (x 2 - 5x + 6) (1) - (x + 6) (2x - 5) = dx (x 2 - 5x + 6)2

dy = 1 at point (0, 1) dx \ Slope of normal = – 1 Now equation of normal is y – 1 = –1 (x – 0) Þy–1 =–x x+y= 1

40. (c)

= –2 < 0 dr 2 \ For r = 5 A is maximum q r From (1) 20 - 2(5) 10 = =2 q = 5 5 2 ´ p(5) 2 = 25 sq. m A= 2p 1 Eccentricity of ellipse = 2 1 a Now, – = – 4 Þ a = 4 × = 2 Þ a = 2 2 e æ 1ö 2 2 2 We have b = a (1 – e ) = a2 ç 1 - ÷ = 4 è 4ø 3 =3 4 \ Equation of ellipse is

×

x 2 y2 + =1 4 3 Now differentiating, we get x 2y ´ y¢ = 0 Þ y¢ = – 3x Þ + 2 3 4y 3 2 1 y¢ (1,3/ 2) = - ´ = 4 3 2 Slope of normal = 2 3 \ Equation of normal at æç1, ö÷ is è 2ø 3 y – = 2 (x – 1) Þ 2y – 3 = 4x – 4 2 \ 4x – 2y = 1

M-131

Integrals

22

Integrals 1.

10 p

ò0

| sin x | dx is (a) 20 (c) 10

In =

ò

(b) 8 (d) 18

2

[2002] (b) 1 (d) zero

]dx is

[2002]

[2003]

0

n®¥

1 2 (c) ¥

ò [x

ò f ( x ) g ( x ) dx, is

tan x dx then lim n[ I n + I n + 2 ] equals

(a) 2

Let f(x) be a function satisfying f '(x) = f(x) with f(0)=1 and g(x) be a function that satisfies f(x) + g(x) = x 2 . Then the value of the integral

n

0

3.

6.

1

p/4

2.

[2002]

7.

(a) e +

e2 5 + 2 2

(b) e -

e2 5 2 2

(c) e +

e2 3 2 2

(d) e -

e2 3 - . 2 2

1

The value of the integral I = ò x(1 - x ) n dx is 0

0

(a) 2 – 4.

(b) 2 +

2

2 –1 (c) p 2 x (1 + sin x )

ò

2 - p 1 + cos x p2 (a) 4

(d) - 2 - 3 + 5

dx is

(c) zero 5.

2

[2002] (b) p 2 (d)

8.

p 2

(a)

(b)

1 n +1

(c)

1 n+2

(d)

1 1 . n +1 n + 2

n

r

1 å n e n is r =1 (a) e + 1 (c) 1 – e Lim n®¥

b

If f (a + b - x) = f ( x) then ò xf ( x )dx is equal to a

b

(a)

a+b f (a + b + x )dx 2 ò

9.

The value of

(c)

a +b b ò f ( x)dx 2 a

(d)

b-a b ò f ( x )dx . 2 a

[2004] (b) e – 1 (d) e

3

ò |1- x

2

|dx is

[2004]

-2

[2003] (a)

1 3

(b)

14 3

(c)

7 3

(d)

28 3

a

a+b b (b) ò f (b - x)dx 2 a

[2003]

1 1 + n +1 n + 2

p/2

10. The value of I =

ò

0

(a) 3 (c) 2

(sin x + cos x ) 2 1 + sin 2 x

(b) 1 (d) 0

dx is

[2004]

EBD_7764 M-132

Mathematics p

11.

If

ò xf (sin x)dx = A

p/2

0

ò

1

f (sin x ) dx, then A is

16.

0

0

[2004] (a) (c) 12.

2

(b) p

2p p 4

If f ( x) =

, I1 =

ò

xg{x (1 - x )}dx

ò

3

g{x (1 - x )}dx, then the value

(c) I3 = I 4

(d) I3 > I 4

17. The value of

ò

cos 2 x

- p 1+ a

x

dx , a > 0, is

f (-a )

13.

14.

(c) (b) –3 (d) 2

(a) (- cos a, sin a)

(b) (cos a, sin a)

(c) (- sin a, cos a)

(d) (sin a, cos a)

(b) (c) (d)

2

[2004]

(c)

19.

ò xf (sin x)dx

0

æ x pö log tan ç - ÷ + C è 2 8ø 2

-

1

1 + x2

is equal to

p

æ x 3p ö log tan ç - ÷ + C è2 8 ø 2

+C

(b)

(a) p ò f (cos x )dx

1

xe x

3 2 (d) 1

1 2 (c) 2

(a)

(c)

+C

dx is

[2006]

æ xö log cot ç ÷ + C è 2ø 2

(log x)2 + 1

x

9- x + x 3

1

log x

p 2

(d) 2p

18. The value of integral, I =ò

æ x 3p ö log tan ç + ÷ + C è2 8 ø

(b) (d)

[2005]

p 2

ò

20.

ïì (log x - 1) ïü ò íï1 + (log x)2 ýï dx is equal to î þ

(a)

p a

[2005]

[2006]

0

2

15.

dx

p

dx ò cos x - sin x is equal to 1

(b)

6

sin x dx = Ax + B log sin( x - a ), +C , If ò sin( x - a ) then value of (A, B) is [2004]

(a)

(a) a p

[2004]

(a) 1 (c) –1

x2

[2005] (b) I1 > I 2

p

I of 2 is I1

ò2 1

(a) I 2 > I1

f (-a )

f ( a)

and I 2 =

0

2

1

f ( a)

1+ ex

3

x ò 2 dx , I3 =

x and I 4 = ò 2 dx then

(d) 0 ex

2

x If I1 = ò 2 dx , I 2 =

1

-

p 2

0

x2 + 1 x

(log x)2 + 1

+C

(d) p

ò

f (cos x )dx

0

3p 2

(a)

+C

f (sin x)dx

[( x + p )3 + cos 2 ( x + 3p )]dx is equal to

[2006] 4

x

0

p /2

p /2

ò

p

(b) p ò f (sin x )dx

(c)

p 32 p 2

4

p p + 32 2 p (d) -1 4

(b)

M-133

Integrals a

21.

ò [ x] f '( x)dx , a > 1 where

The value of

[x]

2ò

26. The value of

1

denotes the greatest integer not exceeding x is [2006] (a) af (a) - { f (1) + f (2) + .............. f ([ a])} (b) [a ] f (a) - { f (1) + f (2) + .............. f ([a ])}

(a)

pö æ x + log | cos ç x – ÷ | + c è 4ø

(b)

pö æ x – log | sin ç x – ÷ | + c è 4ø

(c)

pö æ x + log | sin ç x – ÷ | +c è 4ø

(d)

pö æ x – log | cos ç x - ÷ | + c è 4ø

(c) [a ] f ([a ]) - { f (1) + f (2) + .............. f ( a)} (d) af ([ a]) - { f (1) + f (2) + .............. f ( a )} 22.

dx

ò cos x +

3 sin x

equals

[2007] 27.

(d)

1 æx pö log tan ç - ÷ + C 2 è 2 12 ø

0

1

(1) = 41. Then

x

23.

24.

log t 1 dt , Let F(x) = f (x) + f æç ö÷ ,where f ( x) = ò 1+ t è xø l

Then F(e) equals [2007] (a) 1 (b) 2 (c) 1/2 (d) 0, The solution for x of the equation x

ò

dt 2

2

t t -1

(a)

3 2

=

(b) 41

(c) 42

(d)

29. The value of

ò

Let I = ò

0

sin x x

1

dx and J = ò

cos x

0

x

which one of the following is true?

the value of

ò x éë x 0

dx. Then

[2008]

(a) I >

2 and J > 2 3

(b) I


2 and J < 2 3

1+ x2

dx is

[2011]

p p log 2 log 2 (b) 8 2 (c) log 2 (d) p log 2 Let [.] denote the greatest integer function then 1.5

1

8log(1 + x)

41

(a)

30.

(d) None of these

[2010]

(a) 21 1

[2007]

(b) 2 2

ò p( x) dx equals 0

0

p is 2

(c) 2 25.

ò [cot x] dx , where [ . ] denotes the greatest

integer function, is equal to : [2009] (a) 1 (b) –1 p p (d) (c) 2 2 28. Let p(x) be a function defined on R such that p¢(x) = p¢(1 – x), for all x Î [0, 1], p (0) = 1 and p

(b) log tan çæ x - p ÷ö + C è 2 12 ø 1 x p log tan çæ + ÷ö + C 2 è 2 12 ø

[2008]

p

æx pö (a) log tan ç + ÷ + C è 2 12 ø

(c)

sin xdx is pö æ sin ç x – ÷ è 4ø

(a) 0 (c)

3 4

2ù

û dx is :. 3 2 5 (d) 4

[2011 RS]

(b)

31. If the 5 tan x dx = x + a ln sin x - 2 cos x + k , tan x - 2 then a is equal to : [2012] (a) – 1 (b) – 2 (c) 1 (d) 2

ò

EBD_7764 M-134

32.

Mathematics

x

ò

If g (x) = cos 4t dt , then g (x + p) equals 0

g ( x) g ( p)

(b) g (x) + g (p)

(c) g (x) – g (p)

(d) g (x) . g (p)

(a)

33.

[2012]

If ò f ( x)dx = y(x), then to

òx

5

f ( x 3 )dx is equal [2013]

1

x+ 36. The integral ò æç1 + x - 1 ö÷ e x dx is equal to xø è [2014]

(a) (c)

( x + 1) e

x+

1 x

+c

(b) - xe

( x - 1) e

x+

1 x

+c

(d) xe

p

ò

37. The integral

1 + 4sin 2

0

ò

p/61+

tan x b

Statement-2 :

ò

a

35.

(b) 4 3 - 4 -

(c) p - 4

(d)

38. The integral

is equal to p/6

2p - 4- 4 3 3

equals :

[2015]

1 - (x 4 + 1) 4

æ x 4 + 1ö 4 (b) - ç 4 ÷ + c è x ø

+c

1

b

f ( x )dx = ò f (a + b - x )dx. a

x

ò

dx

ò x 2 (x 4 + 1)3/4

p 3

1

[2013] (a) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. The intercepts on x-axis made by tangents to thecurve, y =

+c

(a) 4 3 - 4

(a)

dx

1 x

+c

[2014]

1é 3 x y ( x 3 ) - ò x 2 y ( x3 )dx ùû + C 3ë 1 3 x y ( x3 ) - 3ò x 3 y ( x 3 )dx + C (b) 3 1 3 x y ( x3 ) - ò x 2 y ( x3 )dx + C (c) 3 1é 3 x y ( x 3 ) - ò x3 y ( x3 ) dx ùû + C (d) 3ë Statement-1 : The value of the integral p /3

x+

1 x

x x - 4sin dx equals: 2 2

(a)

34.

x+

t dt , x Î R, which are parallel

æ x 4 + 1ö 4 (c) ç 4 ÷ + c è x ø 4

39. The integral

log x 2

ò log x 2 + log(36 - 12 x + x 2 ) dx 2

is equal to : (a) 1 (c) 2

(b) 6 (d) 4

40. The integral ò

2x12 + 5x 9

(x

(a)

5

3

x5

(

5

3

[2013] (b)

2

)

- x10 2 x5 + x3 + 1

(

2

)

+C

+C

3

)

+ x +1

2 x + x +1

0

to the line y = 2x, are equal to : (a) ± 1 (b) ± 2 (c) ± 3 (d) ± 4

1

(d) (x 4 + 1) 4 + c

[2015]

dx is equal to :

[2016]

M-135

Integrals

(c)

(d)

-x

(

(a) –1 (c) 2

5

x5 + x3 +1 x10

+C

2

)

2 x5 + x3 +1

n 42. Let In = ò tan x dx,(n > 1) . I4 + I6 = a tan5x + bx5

+ C, where C is constant of integration, then the ordered pair (a, b) is equal to : [2017]

+C

2

( ) where C is an arbitrary constant. 3p 4

ò

41. The integral

p 4

(b) –2 (d) 4

dx is equal to : 1 + cos x

[2017]

æ 1 ö (a) çè - , 0÷ø 5

æ 1 ö (b) çè - ,1÷ø 5

æ1 ö (c) çè , 0÷ø 5

æ1 ö (d) çè , -1÷ø 5

Answer Key 1 2 (a) (b) 16 17 (b) (b) 31 32 (d) (b, c)

3 (d) 18 (b) 33 (c)

10 p

1.

(a)

ò

I=

0p

4 (b) 19 (d) 34 (d)

5 (c) 20 (c) 35 (a)

6 (d) 21 (b) 36 (d)

7 (d) 22 (c) 37 (b)

8 (b) 23 (c) 38 (b)

9 (d) 24 (d) 39 (a)

10 (c) 25 (b) 40 (d)

| sin x | dx = 10 ò | sin x | dx 0

= lim

n®¥

[Q | sin x| is periodic with period p and sin x > 0 if 0 < x < p]

ò

0

p/2

sin x dx = 20 [ - cos x ]0

(b)

I n + I n+ 2 =

ò

n

= 20

3.

(d)

2

1

0

0

2 2 ò éë x ùû dx = ò éë x ùû dx +

ò

2

tan x(1 + tan x )dx

2

=

ò

0

=

p/4

é tan n +1 x ù tan x sec x dx = ê ú êë n + 1 úû 0 n

1- 0 1 = n +1 n + 1

2

15 (d) 30 (c)

n =1 æ 1ö n ç1 + ÷ è nø

0

p/ 4

14 (a) 29 (d)

3

p/4

2.

13 (b) 28 (a)

1 lim n [I + I ] Þ n®¥ n n+2 n +1 1 n = lim n. = lim n ®¥ n + 1 n®¥ n + 1

0

I = 20

12 (d) 27 (c) 42 (c)

\ In + In+2 =

p

= 10ò sin x dx

p/2

11 (b) 26 (c) 41 (c)

1

= ò 0dx + 0

2

ò

ò

1

1

ò

2

2dx +

2

3 2

é x 2 ù dx + ë û

éx2 ù + ë û

3

1dx +

= [ x ]1 + [ 2 x ] 2

2

2

ò

3

ò 3dx 3

+ [ 3x] 3

2

é x 2 ù dx ë û

EBD_7764 M-136

Mathematics [Q given that f(a + b – x) = f(x)]

= 2 -1 + 2 3 - 2 2 + 6 - 3 3

b

2I = (a + b)ò f ( x)dx

= 5- 3 - 2 4.

(b)

2 x (1 + sin x)

p

ò-p

1 + cos 2 x 2 x dx

p

= ò-p

2

1 + cos x

a b

dx Þ

+ 2ò

x sin x

p

-p 1 + cos 2

x

6.

Integrating log f ( x) = x + c Þ f ( x ) = e x + c f (0) = 1 Þ f ( x ) = e x

a

= 2ò f ( x ) dx if f(x) is even.

1

I = 4 ò0 I=4

1 + cos2 (p - x )

p (p - x)

ò0

sin x

1 + cos 2 x

sin x dx

0

1 + cos 2 x

Þ I = 4pò

Þ 2I = 4p ò

=

- 4ò

1 + cos 2 x 7.

dx 1 + cos 2 x put cos x = t Þ - sin x dx = dt 0

-1

\ I = -2 p

1

ò 1+ t2 1

1

dt = 2p ò

(d)

1

-1 1 + t

2

dt

8.

(b)

-1 -1 = 2p éë tan 1 - tan ( -1) ùû

a

b

b

= (a + b )ò f (a + b - x)dx - ò xf (a + b - x )dx a b

a

b

a

a

= (a + b )ò f ( x) dx - ò xf ( x)dx

[ ] - 2[xe

x

- ex

]

1 0

-

[ ]

1 2x e 2

1 0

1

1

0

0

I = ò x(1 - x )n dx = ò (1 - x )(1 - 1 + x) n dx 1

n

r

n

1 å ne n [Using definite integrals as r =1 limit of sum] Lim n®¥

= ò e x dx = e - 1 0

3

b

I = ò xf ( x )dx = ò (a + b - x ) f (a + b - x )dx a

0

1 x 2e x 0

1

é p æ -p ö ù p = 2p ê 4 - ç 4 ÷ ú = 2p. 2 = p2 è ø ë û

(c)

0

é x n +1 x n + 2 ù = ò (1 - x) x dx = ê ú êë n + 1 n + 2 úû 0 0 1 1 = n +1 n + 2

1

5.

0

1

-1 = 2p éë tan t ùû -1

b

1

é e2 1 ù 2 = e - ê - ú - 2[e - e + 1] = e - e - 3 2 2 ëê 2 2 úû

x sin x dx

sin x

p

0 1

= ò x 2 e x dx - ò e 2 x dx

dx

dx

p

1

\ ò f ( x ) g ( x) dx = ò e x ( x 2 - e x ) dx

0

x) sin (p - x )

f ¢( x ) =1 f ( x)

(d) Given f ¢ ( x ) = f ( x) Þ

x sin x dx

p (p -

( a + b) f ( x )dx 2 ò a

dx

é a ù êQ ò f ( x ) dx = 0ú = 0+4 ò0 ; úû 1 + cos2 x êë - a if f(x) is odd p

I=

9.

(d)

ò |1 - x

2

-2

3

| dx =

ò|x

-2 ì 2

2

- 1| dx

x - 1 if ïï 2 Now | x - 1|= í1 - x if ï 2 ïî x - 1 if \ Integral is 2

x £ -1 -1 £ x £ 1 x ³1

M-137

Integrals -1

ò (x

2

-2

1

3

- 1)dx + ò (1 - x2 )dx +ò ( x 2 - 1)dx -1

12. (d)

f ( x) =

1

-1

1

3

1+ e

Þ f (- x ) =

x

e

-x

1 + e- x

1

=

é x3 ù é é x3 ù x3 ù = ê 3 - xú + ê x - 3 ú + ê 3 - xú úû -2 ëê ëê ûú -1 ëê ûú1

e

x

x

e +1 \ f ( x) + f ( - x) = 1 " x f (a )

æ 1 ö æ 8 ö æ27 2 ö 1 = ç - + 1÷ - ç - + 2 ÷++æçç 2 -- 3÷ - æç - 1ö÷ 3 3 è ø è ø èè 3 3 ø è 3 ø

ò

Now I1 =

f (a )

2 2 4 2 28 = + + +6+ = 3 3 3 3 3

10.

(c)

p 2

ò

I=

(sin x + cos x )2 1 + sin 2 x

0

=

p 2

I= ò

0 p 2

ò

= (sin x + cosx)dx 0

pù é êQ sin x + cos x > 0 if 0 < x < 2 ú ë û p 2 0

or I = [ - cos x + sin x ] = 2 p

11.

0

= ò (p - x) f (sin x)dx 0

p 2

\ 2 I = pò f (sin x)dx = p.2 ò f (sin x )dx p 2

0

\ I = p ò f (sin x )dx Þ A = p 0

1 2

æ p x pö log tan ç + + ÷ + C è 4 2 8ø

é æ p xö ù êQ ò sec x dx = log tan çè + ÷ø ú 4 2 û ë 1 æ x 3p ö log tan ç + ÷ + C = è2 8 ø 2

p

2

sin x

ò sin( x - a) dx = ò

=

(b) Let I = ò xf (sin x )dx

p

(1 - x) g{x (1 - x )}dx

sin( x - a + a) dx sin( x - a) sin( x - a ) cos a + cos( x - a)sin a =ò dx sin( x - a ) = ò {cos a + sin a cot( x - a )}dx = (cos a) x + (sin a) logsin( x - a) + C \ A = cosa, B = sin a dx dx 14. (a) ò = pö cos x - sin x ò æ 2 cos ç x + ÷ è 4ø 1 pö æ = ò sec çè x + 4 ÷ø dx 2 13. (b)

(sin x + cos x )2 dx (sin x + cos x )

ò

f ( - a)

b b é ù ê using ò f ( x ) dx a = ò f ( a + b - x ) dx ú êë úû a a = I 2 - I1 Þ 2 I1 = I 2

dx

We know [(sin x + cos x)2 = 1 + sin 2 x] , so

xg{x(1 - x )}dx

f ( - a)

15. (d)

(log x - 1) 2

ò (1 + (log x)2 )2 dx =ò

1 + ( log x ) - 2 log x 2

é1 + ( log x )2 ù ë û

2

dx

é ù 1 2 log x dx = òê 2 2 2ú êë (1 + (log x) ) (1 + (log x) ) úû

EBD_7764 M-138

Mathematics

é et 2t et ù = òê ú dt put log x = t 2 (1 + t 2 )2 ûú ëê1 + t

é Which is of the form e ( f ( x) + f '( x ) ) dxù ò ë û t e x +c = +c = 1+ t2 1 + (log x ) 2

16.

2

0

1

x2

6

3

2

p

17.

p

=

cos

....(1)

( -x)

dx ò ....(2) x -p 1 + a Adding equations (1) and (2) we get p æ 1+ ax ö = cos 2 x dx dx 2 I = ò cos x ç ò x÷ 1 + a è ø -p -p 2

= 2ò cos 2 x dx

0 p

0

p /2 0

20. (c)

I=

= 2 ´ 2 ò cos 2 x dx = 4 ò sin 2 x dx

ò (1 - cos 0

[( x + p )3 + cos 2 ( x + 3p )]dx

3p 2

Put x + p = t p 2

ò [t

3

2

)

x dx

p 2

+ cos 2 t ]dt = 2

ò cos

p 2

2

tdt

[using the property of even and odd function] p 2

0

p 2

ò

p 2

p 2

0

p 2 -

I=

p 2

0

p

= p ò f (cos x ) dx

0

2 Þ I = 2 ò sin x dx = 2

p

p Þ 2 I = p ò f (sin x )dx 0 p /2 pp I = ò f (sin x )dx = p ò f (sin x )dx 20 0

p

p 2

a

I = ò xf (sin x)dx = ò (p - x ) f (sin x)dx

cos 2 x

p

b

f ( x)dx = ò f (a + b - x )dx ]

0

dx ò -x -p 1 + a

p

… (2)

= p ò f (sin x) dx - I

b b é ù ê Using ò f ( x ) dx = ò f ( a + b - x ) dx ú êë úû a a

=

dx

6

19. (d)

cos 2 x

ò 1 + a x dx -p 2

ò

… (1)

3 2 I = ò dx = 3 Þ I = 2 3

3

0

(b) Let I =

b

dx

Adding equation (1) and (2)

3

x x Þ ò 2 dx > ò 2 dx Þ I1 > I 2 0

9-x+ x

a

x3

1

2

9- x

[ using

0

1

9- x + x 3

3

I3 = ò 2 dx, I 4 = ò 2 dx " 0 < x < 1, x > x 0

x

I =ò

I =ò

0

1

0

6

18. (b)

I1 = ò 2 x dx, I 2 = ò 2 x dx,

(b)

0

p æ pö Þ I + I = 2ç ÷ = p Þ I = è 2ø 2

x

1

p 2

2 Þ I = 2 ò dx - 2 ò cos x dx

Þ dx = et dt 2t ù t é 1 = òe ê ú dt 2 (1 + t 2 )2 ûú ëê1 + t

1

p 2

=

p

ò (1 + cos 2t )dt = 2 + 0 0

M-139

Integrals

21.

(b) Let a = k + h where k is an integer such that [a] = k and 0 £ h < 1 a

2

3

1

1

2

dz 1 1 = z Þ - 2 dt = dz Þ dt = – 2 t t z and limit for t = 1 Þ z = 1 and for t = 1/e Þ z=e

\ Put

\ò [ x] f '( x )dx = ò 1 f '( x )dx + ò 2 f '( x )dx+ k +h

k

...

ò

(k - 1)dx +

k -1

22.

ò

kf '( x )dx

\

=ò

e

e

1

log z æ dz ö ç- ÷ ( z + 1) è z ø

F(e) =

=ò

e log t

ò1

1/ e log

ò1

t dt 1+ t

log t

e (log t )(t + 1) + log t dt = ò dt 1 t (1 + t ) t (1 + t )

Þ F(e) =

e log t

ò1

t

dt \

Let log t = x

1 dt = dx t

[for limit t = 1, x = 0 and t = e, x = log e = 1] 1

\ F(e) =

t dt 1+ t

Now for solving, I =

e

e t.log t

x log

1/ e log t t dt + ò dt ....(A) 1 1+ t 1+ t

b

ò1 1 + t dt + ò1 t (1 + t ) dt

1

1 x p . log tan æç + ö÷ + C è 2 12 ø 2 æ 1ö (c) Given F (x) = f (x) + f ç ÷ ,where è xø

e log

b

òa f (t )dt = òa f ( x)dx ]

Equation (A) becomes

I=

\ F(e) = f (e) + f æç 1 ö÷ è eø

[Q log1 = 0]

log z dz z ( z + 1)

[By property

ò cosec x dx = log | (tan x / 2) | + C

ò1

æ dz ö çè - 2 ÷ø z

e log t dt \ I= ò 1 t (t + 1)

1 pö æ . cosec ç x + ÷ dx è 2ò 6ø But we know that

Þ F(e) =

z +1

1

=ò

æ dz ö çè - 2 ÷ø z

e (log1 - log z ).z

=ò -

Þ I=

f (x) =

æ 1ö log ç ÷ è zø 1 1+ z

1

1 dx = ò p 2 é p ù êësin 6 cos x + cos 6 sin x úû 1 dx = .ò pö 2 æ sin ç x + ÷ è 6ø

23.

ò1

I=

k

= {f (2) – f (1)} + 2{f (3) – f (2)} + 3{f (4) –f (3)}+ ........ + (k – 1) {f(k) – f(k – 1)} + k{f(k + h) – f(k)} = – f (1) – f (2) – f (3) ......... – f (k) + kf (k + h) = [a] f (a) - { f (1) + f (2) + f (3) + ¼ f ([a])} dx (c) I = ò cos x + 3 sin x dx Þ I =ò é1 ù 3 2 ê cos x + sin x ú 2 ë2 û

\

e

ò0 x dx

Þ F(e) =

24. (d)

x

ò 2t

1

1 2

dt 2

t -1

\ ésec-1 t ù ë

= x

û 2

p 2 =

p 2

é x2 ù F(e) = ê ú ëê 2 ûú 0

EBD_7764 M-140

Mathematics é ù dx = sec -1 x ú êQ x x2 - 1 ëê ûú

ò

Þ

sin x 1

Þ

ò

Þ

ò

Þ

1

x

x

ò

ò

cos x


is 4 4 p æpö ö æ ç b sin b + cos b + 2b ÷. Then f ç ÷ [2005] 4 è2ø ø è ö æp (a) ç + 2 - 1÷ (b) ø è4

ö æp ç - 2 + 1÷ ø è4

ö æ p ö æ p (c) ç1 - - 2 ÷ (d) ç1 - + 2 ÷ ø è 4 ø è 4 8. The area enclosed between the curves y2 = x and y = | x | is [2007] (a) 1/6 (b) 1/3 (c) 2/3 (d) 1 9. The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1is equal to [2008] 1 5 (a) (b) 3 3 4 (c) 2 (d) 3 3 10. The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is: [2009] (a) 6 (b) 9 (c) 12 (d) 3 11. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 3p is [2010] 2

(a) 4 2 + 2

(b) 4 2 - 1

(c) 4 2 + 1

(d) 4 2 - 2

M-145

Applications of Integrals

12.

The area of the region enclosed by the curves

13.

1 y = x, x = e, y = and the positive x-axis is [2011] x 3 (a) 1 square unit (b) square units 2 1 5 (c) square units (d) square unit 2 2 The area bounded by the curves [2011 RS]

y 2 = 4 x and x2 = 4y is:

14.

32 sq units 3

(b)

(c)

8 sq. units 3

(d) 0 sq. units

y and 4 x2 = 9y and the straight line y = 2 is : [2012]

The area between the parabolas x 2 =

(b)

(a)

}

+ y 2 £ 1 and y 2 £ 1 - x is:

p 2 2 3

(b)

p 2 + 2 3

p 4 p 4 + (d) 2 3 2 3 17. The area (in sq. units) of the region described by {(x, y) : y2 £ 2x and y ³ 4x – 1} is [2015]

(c)

(a)

15 64

(b)

9 32

7 5 (d) 32 64 18. The area (in sq. units) of the region {(x, y) : y2 ³ 2x and x2 + y2 £ 4x, x ³ 0, y ³ 0} is : [2016]

(a) p-

y = x , 2y – x + 3 = 0, x-axis, and lying in the first quadrant is : [2013] (a) 9 (b) 36 (d)

2

[2014]

10 2 3

20 2 (d) 10 2 3 The area (in square units) bounded by the curves

(c) 18

{( x, y ) : x

(c)

(c)

15.

A=

16 sq units 3

(a)

(a) 20 2

16. The area of the region described by

4 2 3

(b)

p 2 2 2 3

4 8 (d) p3 3 19. The area (in sq. units) of the region

(c) p-

{(x, y) : x ³ 0, x + y £ 3, x2 £ 4y and y £ 1 + x } is : [2017]

27 4

(a)

5 2

(b)

59 12

(c)

3 2

(d)

7 3

Answer Key 1 (d) 16 (c)

2 (a) 17 (b)

3 (d) 18 (d)

4 (d) 19 (a)

5 (a)

6 (d)

7 (d)

8 (a)

9 (d)

10 (b)

11 (d)

12 (b)

13 (b)

14 (c)

15 (a)

EBD_7764 M-146

Mathematics

2

1.

(d) We have

ò

0

3.

3 f ( x)dx = ; Now,, 4

2

2

2

0

0

0

(d) (0,3)

y=–x+1

y=x–1

(–1,2)

(2,1)

ò xf ¢( x)dx = xò f ¢( x)dx - ò f ( x)dx 2 = [ x f ( x)]0 -

2.

3 3 = 2 f (2) 4 4

(1,0)

(–1,0)

(1,0)

1

(1,0)

(–1,0)

(1,0)

0

=

x

0

1

-1

0

1 2

ò (2 + 2 x)dx + ò 2dx + ò (4 - 2 x)dx

0 é2 x - x 2 ù ë û -1

1

2

+ [ 2 x] + é4 x - x2 ù ë û1 = 0 - ( -2 + 1) + (2 - 0) + (8 - 4) - (4 - 1) 1 0

= 1 + 2 + 4 - 3 = 4 sq. units

x

Note: Graph of y = | f(x) | can be obtained from the graph of the curve y = f(x) by drawing the mirror image of the portion of the graph below x-axis, with respect to x-axis. Clearly the bounded area is as shown in the following figure. y

4.

(d) The required area is shown by shaded region Y (y = 2 – x)

(y = x – 2)

y = –ln x

y = –ln (–x) –1

2

ò { (3 - x) - (- x + 1)} dx + ò { (3 - x) - ( x - 1)} dx

y=|l n |x|| x

ò { (3 + x) - (- x + 1)} dx +

-1

=

y=|l n x|

y=3–x

0

A=

3 3 = 0 - (Q f (2) = 0) = - . 4 4 (a) First we draw each curve as separate graph y=ln|x| y=lnx x

(1,0)

y=3+x

O

y = ln (–x)

x

1 y = ln x

1

Required area = 4 ò (- lnx )dx 0

= - 4 [ x l n x - x ]0 = 4 sq. units 1

3

X

Required Area 3

3

1

2

A = ò | x - 2 |dx = 2ò ( x - 2)dx 3

é x2 ù = 2 ê - 2xú = 1 ëê 2 ûú 2

M-147

Applications of Integrals

5.

(a) The graph of the curve y = log e ( x + e ) is as shown in the fig.

4

(1–e, 0)

y

y=loge(x+e) (0,1) x

0

é 3 ù ê 2 x 2 x3 ú x - ú Also S2 = ò ç 2 x - ÷ dx = ê 4ø 12 ú ê 3 0è êë 2 úû 0 4 16 16 sq. units = ´8 = 3 3 3 \ S1 : S2 : S3 = 1:1:1 (d) Given that 4æ

7.

2ö

b

ò

x+e=0

p/4

Required area A=

ydx =

ò

1-e

ò

log e ( x + e)dx

f ( b ) = b cos b + sin b –

1- e

put x + e = t Þ dx = dt also At x = 1 – e, t=1 At x = 0, t = e e

\ A = ò loge tdt = [ t log e t - t ]1

e

8.

1

e - e - 0 +1 = 1 Hence the required area is 1 square unit.

6.

Y

(d) Inter section points of x = 4 y and

y = -x

y 2 = 4 x are (0, 0) and (4, 4). The graph is as shown in the figure. y 2 x =4y

(0, 0) 0

(4, 4) S3 x=4

p sin b + 4

y=x

(1, 1) A y1

X'

y2 = x

y2

(1, 0)

(0, 0) O

X

y2=4x Y'

x

\ Required area =

1

ò0 ( y2 - y1 )dx 1

é x3 / 2 x 2 ù - ú = ò ( x - x)dx = ê 2 úû 0 ëê 3 / 2 0 1 1 2 \ Required area = é x3 / 2 ù - 1 é x 2 ù ë û ë û 0 2 0 3 2 1 1 = - = 3 2 6 x 2 2 (d) x + 2 y = 0 Þ y = – 2 [Left handed parabola with vertex at (0, 0)] 1 x + 3 y 2 = 1 Þ y 2 = – ( x – 1) 3 [Left handed parabola with vertex at (1, 0)] 1

By symmetry, we observe 4

S1 = S3 =

ò ydx 0

4 2

=ò

0

x dx = 4

2

p p æ pö æ p ö f ç ÷ = ç1 - ÷ sin + 2 = 1 - + 2 è 2ø è 4ø 2 4 (a) The area enclosed between the curves y2 = x and y = | x | From the figure, area lies between y2 = x and y = x

2

y=4 S 1 S2

p cos b + 2b 4

Differentiating w. r . t b

0

0

f ( x )dx = b sin b +

4

é x3 ù 16 sq. units ê ú = 3 ëê 12 ûú0

9.

EBD_7764 M-148

Mathematics

Solving the two equations we get the points of intersection as (–2, 1), (–2, –1)

3

2 = 3 + ò ( y - 2) + 1 dy 0

Y

3

é ( y - 2)3 ù = 3+ ê + yú úû 0 ëê 3

A (–2, 1)

D (1, 0)

C

X´

X

11.

B

(d)

8ù é1 = 3 + ê + 3 + ú = 3 + 6 = 9 Sq. units 3û ë3

(–2, –1)

cos x

Y´

sin x

The required area is ACBDA, given by

p

3 ù1

1

0

é 5y 2 2 = ò (1 – 3 y – 2 y )dy = ê y – 3 ú ëê ûú –1 –1 æ 5ö = çè1 – 3 ÷ø

10.

5ö æ – ç –1 + ÷ = 2 ´ 2 = 4 sq. units. è 3ø 3 3

(b) The given parabola is (y – 2)2 = x – 1 Vertex (1, 2) and it meets x –axis at (5, 0) Also it gives y2 – 4y – x + 5 = 0 So, that equation of tangent to the parabola at (2, 3) is 1 y.3 – 2 (y + 3) – (x + 2) + 5 = 0 2 or x – 2y + 4 = 0 which meets x-axis at (– 4, 0). In the figure shaded area is the required area. Let us draw PD perpendicular to y – axis. Y D

P

(0,2)A

p

4

2

3p

4

2

2p

Area above x-axis = Area below x-axis \ Required area p é ù 2 ê ú ê p (cos x - sin x)dx + ò (sin x - cos x)dx + ú ê4 ú p ê ú 4 = 2êò ú p ê0 ú sin x dx ú ê ò ê ú p êë úû 2

= 4 2 -2 12. (b) Area of required region AOBC 1

e

1 = ò xdx + ò dx x 0 1 =

(2, 3)

p

5p

1 3 + 1 = sq. units 2 2 y

x=e

y=x

B (–4, 0)

O

X C(5,0)

(1, 1) A

Then required area = Ar DBOA + Ar (OCPD) – Ar (DAPD) 1 = ×4×2+ 2

1 xdy - ´ 2 ´ 1 0 2

ò

3

O

1 B e, e C

x

M-149

Applications of Integrals

13.

(b)

2

3 3 é2 1 2 2 ùú 2 ê = 2 ´ 3.y - ´ .y 2 3 ê3 ú ë û0

é 3 1 3ù 5 = 2 ê2y 2 - y 2 ú = 2 ´ y ê 3 ú 3 ë û

3 2 2 0

5 20 2 = 2. 2 2 = 3 3 15. (a) Given curves are y= x and 2y – x + 3 = 0 On solving both we get y = –1, 3

4 æ x2 ö Area = ò ç 2 x - ÷ dx 4ø 0è

…(1) …(2)

Y 4

é æ x3/ 2 ö x3 ù 4 64 = ê2 ç ÷ - 12 ú = ´ 8 3 / 2 ø 3 12 ëê è ûú 0

32 16 - 16 = sq. units 3 3 y 2 (c) Given curves x = and x2 = 9y are the 4 parabolas whose equations can be written 1 as y = 4x2 and y = x2 . 9 Y 1 2 y = 4x2 y = 9 x y=2

X

=

14.

X

Also, given y = 2. Now, shaded portion shows the required area which is symmetric. 2

\

æ yö Area = 2 çç 9y ÷ dy 4 ÷ø è 0

3

Required area =

0

2

ò

2

1 1 ù é +1 +1 ê 3.y 2 1 y2 ú = 2ê - . ú ê 1 +1 2 1 +1ú 2 ëê 2 ûú 0

2

} dy

3

y3 = 9. 3 0 16. (c) Given curves are x2 + y2 = 1 and y2 = 1 –x. Intersecting points are x = 0, 1 Area of shaded portion is the required area. So, Required Area = Area of semi-circle + Area bounded by parabola

= y2 + 3y –

1

pr 2 + 2 ò 1 - xdx = 2 0

ò

æ yö Area = 2 ç 3 y ÷ dy ç 2 ÷ø è 0

ò { (2 y + 3) - y

1

=

p + 2 ò 1 - x dx (Q radius of circle = 1) 2 0

1

é (1 - x ) 3 2 ù p ú = + 2ê ê -3 ú 2 2 û0 ë

=

p 4 p 4 - (-1) = + Sq. unit 2 3 2 3

EBD_7764 M-150

17.

(b)

Mathematics Required area

Points of intersection of the two curves are (0, 0), (2, 2) and (2, –2) Area = Area (OAB) – area under parabola (0 to 2)

B æ 1 ,1ö çè ÷ 2 ø O

1

=

ò

-1/ 2

1

ò

-1/ 2

p ´ (2) 2 - ò 2 x dx = p - 8 4 3 0

C æ 1 - 1ö çè , ÷ø 8 2

y +1 dy 4

2

=

19. (a) x+y=3

1

1

1 é 3 3 ù 9 15 9 27 9 + = = = 4 êë 2 8 úû 48 32 48 96 32 A (2, 2)

18.

(d)

O

4y = x 2

(0, 1)

ù 1 é y2 1 é y3 ù + yú - ê ú = ê 4 ëê 2 úû -1/ 2 2 ëê 3 ûú -1/ 2 =

y=1+ x

(1, 2)

y2 dy 2

B

(2, –2)

(2, 1)

(0, 0)

(1, 0)

(2, 0)

Area of shaded region 1

2

2

0

1

0

x2 (1 x )dx (3 x)dx + + ò ò 4 dx =ò

= x] + 1 0

3 ù1 x2 ú

+ 3x ] 3 úú 2 úû 0

2 1

2

2

x2 ù x3 ù 5 - ú - ú = sq.units Sq.unit 2 úû 12 úû 2 1 0

M-151

Differential Equations

24

Differential Equations 1.

The order and degree of the differential

dy ö æ equation ç1 + 3 ÷ è dx ø

2/3

=4

2 (a) (1, ) 3 (c) (3, 3)

d3y dx3

are

x 2 + y 2 - 2ay = 0, where a is an arbitrary constant is [2004]

[2002]

(b) (3, 1)

(b) 2( x 2 + y 2 ) y ¢ = xy

(d) (1, 2)

The solution of the equation

d y dx 2

(c) ( x 2 - y 2 ) y ¢ = 2 xy = e -2 x

[2002] e -2 x (a) 4

6.

1 -4 x 1 -2 x e + cx + d e + cx 2 + d (d) 4 4 The degree and order of the differential equation of the family of all parabolas whose axis is x - axis, are respectively. (a) 2, 3 (b) 2, 1 [2003] (c) 1, 2 (d) 3, 2. The solution of the differential equation

4.

(1 + y 2 ) + ( x - e tan

(a) xe 2 tan

-1

y

-1

= e tan

(b) ( x - 2) = ke 2 tan (c) 2 xe tan (d) xe tan

-1

-1

y

y

y

)

-1

-1

= e 2 tan

dy = 0 , is dx y

+k

y -1

y

+k

= tan -1 y + k

(d) 2( x 2 - y 2 ) y ¢ = xy Solution of the differential equation ydx + ( x + x 2 y )dy = 0 is

e -2 x (b) + cx + d 4

(a) log y = Cx

(c)

3.

The differential equation for the family of circle

(a) ( x 2 + y 2 ) y ¢ = 2 xy

2

2.

5.

(b) -

[2004]

1 + log y = C xy

1 1 (d) - = C + log y = C xy xy The differential equation representing the family (c)

7.

of curves y 2 = 2c ( x + c ) , where c > 0, is a parameter, is of order and degree as follows : [2005] (a) order 1, degree 2 (b) order 1, degree 1 (c) order 1, degree 3 (d) order 2, degree 2

[2003] 8.

dy = y (log y – log x + 1), then the solution dx of the equation is [2005]

If x

æ xö (a) y log ç ÷ = cx è yø

æ yö (b) x log ç ÷ = cy è xø

æ yö (c) log ç ÷ = cx è xø

æ xö (d) log ç ÷ = cy è yø

EBD_7764 M-152

9.

Mathematics

The differential equation whose solution is 2

dV (t ) = - k (T - t ), dt where k > 0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is [2011]

by differential equation

2

Ax + By = 1 where A and B are arbitrary

10.

constants is of [2006] (a) second order and second degree (b) first order and second degree (c) first order and first degree (d) second order and first degree The differential equation of all circles passing through the origin and having their centres on the x-axis is [2007] (a)

y 2 = x 2 + 2 xy

(a) I -

12.

16.

dy dy (b) 2 y = x 2 - 2 xy dx dx

15.

1 k Consider the differential equation [2011RS] 2 (d) T -

given by : 1 y

2 e y e

1 y

1 e + y e

dy x + y satisfying the condition y(1) =1 is = dy x

(a) 4 -

[2008] (a) y = ln x + x (b) y = x ln x + x2 (c) y = xe(x – 1) (d) y = x ln x + x The differential equation which represents the

(c) 1 + 1 - e (d) 1 - 1 + e y e y e The population p (t) at time t of a certain mouse

c2 x ,

(b) 3 -

1 y

17.

1 y

species satisfies the differential equation

where c1, and c2

dp ( t ) = dt

are arbitrary constants, is [2009] (a) y" = y'y (b) yy" = y' (c) yy" = (y')2 (d) y' = y2 Solution of the differential equation

0.5 p(t) – 450. If p (0) = 850, then the time at which the population becomes zero is : [2012] (a) 2ln 18 (b) ln 9

p cos x dy = y (sin x - y ) dx, 0 < x < is[2010] 2

(c)

(a) y sec x = tan x + c (b) y tan x = sec x + c (c) tan x = (sec x + c) y (d) sec x = (tan x + c) y 14.

k (T - t )2 2

æ 1ö y 2 dx + ç x - ÷ dy = 0. If y (1) = 1, then x is è yø

dy dy (d) x 2 = y 2 + 3 xy . dx dx The solution of the differential equation

family of curves y = c1e

13.

(b) I -

(c) e– kT

(c) x 2 = y 2 + xy

11.

kT 2 2

If

18.

dy = y + 3 > 0 and y (0) = 2, then y (ln 2) is dx

equal to : [2011] (a) 5 (b) 13 (c) – 2 (d) 7 Let I be the purchase value of an equipment and V (t) be the value after it has been used for t years. The value V(t) depreciates at a rate given

1 ln 18 2

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by

19.

(d) ln 18

dP = 100 – 12 x . If the firm dx

employs 25 more workers, then the new level of production of items is [2013] (a) 2500 (b) 3000 (c) 3500 (d) 4500 Let the population of rabbits surviving at time t be gover ned by the differen tial

M-153

Differential Equations

equation

dp ( t ) dt

=

1 p ( t ) –200.If p(0) = 100, 2

then p(t) equals:

20.

(a) 2 (b) 2e (c) e (d) 0 If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy)

21.

[2014]

(a) 600 - 500 e t 2

(b) 400 - 300 e -t 2

(c) 400 - 300 et 2

(d) 300 - 200 e -t 2

æ 1ö dx = x dy, then f ç ç- ÷ ÷ is equal to : è 2ø

Let y(x) be the solution of the differential equation (x log x)

(a)

dy + y = 2x log x, (x ³ 1). dx

Then y (e) is equal to:

2 5

(c) -

[2015]

2 5

[2016]

(b)

4 5

(d)

-

4 5

Answer Key 1 (c) 16 (c)

1.

(c)

2 (b) 17 (a)

3 (c) 18 (c)

4 (c) 19 (c)

5 (c) 20 (a)

2 æ 4d3yö æ d yö 1 3 + = ç ÷ çè d x ÷ø è d x3 ø

2

2.

(b)

4.

(c)

8 (c)

9 (d)

3

3

x( e

x( e

= e -2 x ; dy = e +c; 2 dx -2 dx e

11 (d)

+ cx + d

5.

y 2 = 4a( x - h), 2 yy1 = 4a Þ yy1 = 2a

Differentiating, Þ y12 + yy2 = 0 Degree = 1, order = 2. -1 dy (c) (1 + y 2 ) + ( x - e tan y ) =0 dx -1

dx x e tan y Þ + = dy (1 + y 2 ) (1 + y 2 )

ò

(c)

12 (c)

1 (1+ y 2 )

tan -1 y

tan -1 y

\ 2 xe tan

-2 x

4

10 (a)

I .F = e

-2 x

d y

y= 3.

7 (c)

3

æ d yö æ d yö Þ ç1 + 3 = 16 ç ÷ ÷ è d xø è d x3 ø 2

6 (b) 21 (b)

)=ò

13 (d)

dy

=e e tan

e 2 tan )= 2 -1

y

y

2

-1

y

= e2 tan

e tan

-1

y

dy

+C

-1

x 2 + y 2 - 2ay = 0 Differentiate, 2x + 2 y

15 (a)

tan -1 y

-1

1+ y

14 (d)

y

+k

...........(1)

x + yy ¢ dy dy - 2a =0 Þa= y¢ dx dx

æ x + yy ¢ ö Put in (1), x 2 + y 2 - 2 ç y=0 è y ¢ ÷ø Þ ( x 2 + y 2 ) y '- 2 xy - 2 y 2 y ' = 0 Þ ( x 2 - y 2 ) y ' = 2 xy

EBD_7764 M-154

6.

Mathematics

ln z = ln x + ln c

(b) ydx + ( x + x 2 y )dy = 0

dx x dx x Þ = - - x2 Þ + = - x2 , dy y dy y It is Bernoulli form. Divide by x2 x -2

æ yö x = cx or log v = cx or log ç ÷ = cx. è xø 9.

(d)

æ 1ö dx + x -1 ç ÷ = -1. dy è yø

put x -1 = t , - x -2

= e - log y = y -1

1 1 1 Þ . = log y + C Þ log y =C x y xy

y 2 = 2c ( x + c )

........ (i)

2 yy ' = 2c.1 or yy ' = c

........ (ii)

1 dz dx dv = dz Þ = v z x

2

dy + 2g = 0 Þ g = – dx \ equation (i) be

dy ö æ çè x + y ÷ø dx

ì æ dy ö ü x2 + y2 + 2 í - ç x + y ÷ ý . x = 0 dx ø þ î è Þ x2 + y2 – 2x2 – 2x

xdy = y (log y – log x + 1) dx dy y æ æ yö ö = ç log ç ÷ + 1÷ è xø ø dx x è Put y = vx dy xdv xdv =v+ Þ v+ = v (log v + 1) dx dx dx dv dx xdv = v log v Þ = v log v x dx Put log v = z

d2y

2x + 2y .

( y - 2 xy ')2 = 4 yy '3 ........ (iii) Hence equation (iii) is of order 1 and degree 3.

(c)

....(3)

dy æ dy ö + xç ÷ - y =0 2 è ø dx dx dx Which is a DE of order 2 and degree 1. 10. (a) General equation of circles passing through origin and having their centres on the x-axis is x2 + y2 + 2gx = 0...(i) On differentiating w.r.t x, we get

Þ y 2 = 2 yy ' ( x + yy ') [On putting value of c from (ii) in (i)] On simplifying, we get

8.

2

d2y

2 ìï d2y dy æ dy ö üï x í - By 2 - B ç ÷ ý + By =0 è ø dx ïþ dx dx ïî Dividing both sides by –B, we get

xy

\ Solution is t ( y -1 ) = ò ( y -1 )dy + C

(c)

....(2)

A + By

dt æ 1 ö æ 1ö dt t =1 + t ç ÷ = -1 Þ dy çè y ÷ø dy è y ø It is linear in t. Integrating factor

7.

dy =0 dx

æ dy ö + Bç ÷ = 0 2 è dx ø dx From (2) and (3)

dx dt = dy dy

-

=e

...(1)

Ax + By

We get,

1 ò - dy y

Ax 2 + By 2 = 1

Þ y2 = x2 + 2xy

11.

(d)

dy .y = 0 dx

dy dx

dy x + y y = = 1+ dx x x

Putting y = vx and we get

dy dv =v+x dx dx

dv dx = 1+ v Þ ò = dv x ò dx Þ v = ln x + c Þ y = xlnx + cx As y(1) = 1 \ c = 1 So solution is y = x lnx + x v+x

M-155

Differential Equations

12.

(c) We have y = c1e

(d)

k k V (T ) = I + (T 2 - 2T 2 ) = I - T 2 2 2

Þ

y ¢ = c1c2 e c2 x = c2 y

Þ

2 y ¢¢ = c2 Þ y ¢¢ y - ( y ¢ ) = 0 y y2

Þ

13.

c2 x

y ¢¢ y = ( y ¢ )

dy = y tan x - y 2 sec x dx

1 dy 1 - tan x = - sec x y 2 dx y

...(i)

1 1 dy dt =tÞ= y y 2 dx dx

From equation (i) -

I.F. = e

ò tan x dx

= (e )

log|sec x|

sec x

Solution : t(I.F) = ò (I.F) sec x dx 1 Þ y sec x = tan x + c

14.

(d)

dy = y+3 Þ dx

dy

ò y + 3 = ò dx

Þ ln y +3 = x+ c

Since y (0) = 2, \ l n 5 = c Þ ln y + 3 = x + ln5 When x = ln 2 , then ln |y + 3| = ln 2 + ln5 Þ ln | y + 3 | = ln 10 \ y + 3 = ±10 Þ y = 7, – 13 15.

(a)

dV (t ) = - k (T - t ) dt Þ

ò dVt = - k ò (T - t )dt

V (t ) =

k (T - t )2 +c 2

k at t = 0, V (t) = I Þ V (t ) = I + (t 2 - 2tT ) 2

y2

-

dy

=e

1 y

=ò

x.e

Let

-1 =t y 1

–

1 y

1

e

y3

–

1 y

dy

dy = dt

Þ

y2

Þ

I = - ò tet dt = et - tet

Þ

dt + (tan x )t = sec x dx

1

ò

So

=e

dt - t tan x = - sec x dx

Þ

dx x 1 + = dy y 2 y 3 I.F. = e

2

cos x dy = y (sin x - y ) dx

Let

16. (c)

-

1 y

1

+

xe

1 -y e +c y -

1 y

=e

-

1 y

1

1 + e y +c y

1 + c.e1/ y y Since y (1) = 1 1 c=\ e 1 1 Þ x = 1 + - .e1/ y y e 17. (a) Given differential equation is dp ( t ) = 0.5 p ( t ) - 450 dt Þ x = 1+

Þ

dp ( t ) 1 = p ( t ) - 450 dt 2

Þ

dp ( t ) p ( t ) - 900 = dt 2

Þ

2

dp ( t ) = - éë900 - p ( t ) ùû dt

dp ( t ) = -dt 900 - p ( t ) Integrate both the side, we get

Þ

2

-2

dp ( t )

ò 900 - p ( t ) = ò dt

EBD_7764 M-156

Mathematics

Þ \ 2

ò

Þ

Þ Þ

Let 900 – p (t) = u – dp (t) = du We have, du = dt Þ 2 ln u = t + c u 2ln [900 – p(t)] = t + c when t = 0, p (0) = 850 2 ln (50) = c

ò

900 - p ( t )

t1 = 900 - 50 e 2

ò dP = ò (100 - 12

dP = 100 - 12 x dx

x ) dx

P = 100x – 8x3/2 + C Given when x = 0 then P = 2000 Þ C = 2000 Now when x = 25 then P = 100 × 25 – 8 × (25)3/2 + 2000 = 4500 – 1000 Þ P = 3500 (c) Given differential equation is dp (t ) 1 = p (t ) - 200 dt 2 By separating the variable, we get

é1 ù dp(t) = ê p(t ) - 200ú dt ë2 û dp(t )

d ( p (t )) p (t ) - 200

= ò dt

1 dp(t ) p (t ) - 200 = s Þ = ds 2 2 d p (t )

òæ1

ö ç p (t ) - 200 ÷ è2 ø

= ò dt

2ds = ò dt s 2 log s = t + c

ò

æ p (t ) ö - 200÷ = t + c Þ 2log ç è 2 ø

\ t1 = 2ln 18

(c) Given, Rate of change is

Þ

So,

Þ

Þ dP = (100 – 12 x ) dx By integrating

19.

Let

Þ

t = 50 e 2

Þ p (t ) let p (t1) = 0

18.

2

é æ 900 - p ( t ) ö ù 2 êln ç ÷ú = t 50 øû ë è

t = 900 - 50e 2

0

ò1

= dt 1 p (t ) - 200 2 Integrate on both the sides,

1

p(t ) - 200 = e 2 k 2 Using given condition p(t) = 400 – 300 et/2

Þ

20. (a) Given,

dy æ 1 ö + y=2 dx çè x log x ÷ø 1

I.F. = ò x log x dx e = elog(log x) = log x y. logx = ò 2 log xdx + c y logx = 2[x log x – x] + c Put x = 1, y.0 = –2 + c c=2 Put x = e y loge = 2e(log e – 1) + c y(e) = c = 2 21. (b) y (1 + xy)dx = xdy xdy - ydx = xdx y2 æx ö Þò- d çç ÷÷ =òxdx èyø x x2 - = + C as y(1) = –1 Þ C = 1 y 2 2 æ -1 ö 4 -2x Þfç Hence, y = 2 ç ÷ ÷= è2ø 5 x +1

M-157

Vector Algebra

25

Vector Algebra 1.

r r r If | a |= 4,| b |= 2 and the angle between a and r r r b is p /6 then (a ´ b )2 is equal to [2002] (a) 48 (b) 16 (c)

2.

®

®

a

(a) 16 (c) 4

®

®

®

| a |= 7,| b |= 5,| c |= 3 then ®

4.

7.

If a , b , c are vectors such that a + b + c = 0 and

®

® ® ®

a+ b+ c =0 (a) 25 (c) –25

5.

8. angle

between vector b and c is [2002] (a) 60° (b) 30° (c) 45° (d) 90° r r r If | a |= 5, | b |= 4, | c |= 3 thus what will be the rr rr rr value of | a.b + b .c + c.a | , given that

9.

[2002] (b) 50 (d) –50

r r If the vectors c , a = xiˆ + yjˆ + zkˆ and bˆ = ˆj r r are such that a, c and form a right handed r system then c is: [2002] r ˆ ˆ (a) zi - xk (b) 0 ˆ (c) yj (d) - ziˆ + xkˆ

Ù

Ù

b = 6 i + 3 j are two

®

c is a vector such that ®

34 : 45 : 39

(c) 34 : 39 : 45

® ® ®

® ® ®

®

an d

® ®

(a)

[2002]

(b) 64 (d) 8

Ù

®

®

c = a ´ b then | a |:| b |:| c | [2002]

If a , b , c are vectors such that [ a b c ] = 4 ® ®® ® ® ®

Ù

a = 3 i-5 j

®

®®®

® ®

®

vectors and

(d) none of these

then [ a ´ b b ´ c c ´ a ] =

3.

6.

10.

® ®

® ®

(b)

34 : 45 : 39

(d) 39 : 35 : 34 ® ®

® ® ®

If a ´ b = b ´ c = c ´ a then a + b + c = [2002] (a) abc (b) –1 (c) 0 (d) 2 The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are [2002] (a) 13, 5 (b) 12, 6 (c) 14, 4 (d) 11, 7 A bead of weight w can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle q with the vertical then tension of the thread and reaction of the wire on the bead are (a) T = w cos q R = w tan q [2002] (b) T = 2w cos q R=w (c) T = w R = w sin q (d) T = w sin q R = wcot q

r r r L e t u = iˆ + ˆj, v = iˆ - ˆj and w = iˆ + 2 ˆj + 3kˆ . r If nˆ is a unit vector such that u.nˆ = 0 and r r v .nˆ = 0 , then w. nˆ is equal to [2003] (a) 3 (b) 0 (c) 1 (d) 2.

EBD_7764 M-158

11.

A particle acted on by constant forces

17.

4iˆ + ˆj - 3kˆ and 3iˆ + ˆj - kˆ is displaced from

the point ˆi + 2jˆ - 3kˆ to the point 5iˆ + 4jˆ + kˆ . The total work done by the forces is [2003] (a) 50 units (b) 20 units (c) 30 units (d) 40 units. 12.

The vectors AB = 3iˆ + 4kˆ & AC = 5iˆ - 2 ˆj + 4kˆ are the sides of a triangle ABC. The length of the median through A is [2003] (a)

13.

14.

(b)

288

equal to [2003] (a) 1 (b) 0 (c) –7 (d) 7 A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1) B(2, 1, 3) and C(-1, 1, 2). Then the angle between the faces OAB and ABC will be [2003]

19.

Then P 2 : Q 2 : R 2 is

æ 19 ö ÷ è 35 ø

-1 (b) cos ç

(c) cos -1æç

17 ö ÷ è 31 ø

16.

18

(c) 72 (d) 33 r r r r r r a , b , c are 3 vectors, such that a + b + c = 0 , r r rr rr rr r a = 1, b = 2, c = 3, then a.b + b .c + c .a is

(a) 90 o

15.

18.

(d) 30 o

20.

a

a2

1 + a3

If b

b2

1 + b 3 = 0 and vectors (1, a, a 2 ),

c

c2

1 + c3

and

5iˆ - ˆj + 5kˆ respectively. Then ABCD is a [2003]

(a) parallelogram but not a rhombus (b) square (c) rhombus (d) rectangle.

(c) 21.

[2003]

(a) 2 : 3 : 1 (b) 3 : 1 : 1 (c) 2 : 3 : 2 (d) 1 : 2 : 3. A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by [2003] (a)

(1, b, b2 ) and (1, c, c 2 ) are non- coplanar, then the product abc equals [2003] (a) 0 (b) 2 (c) –1 (d) 1 Consider points A, B, C and D with position vectors 7iˆ - 4 ˆj + 7 kˆ, iˆ - 6 ˆj + 10kˆ , - iˆ - 3 ˆj + 4 kˆ

Mathematics r r r If u , v and w are three non- coplanar vectors, r r r r r r r then (u + v - w).(u - v ) ´ (v - w) equals [2003] rr r (a) 3u .v ´ w (b) 0 rr r r r r (c) u .(v ´ w) (d) u.w ´ v . r A couple is of moment G and the force forming r r the couple is P . If P is turned through a right angle the moment of the couple thus r r formed is H . If instead , the force P are turned through an angle a , then the moment of couple becomes [2003] r r (a) H sin a - G cos a r r (b) G sin a - H cos a r r (c) H sin a + G cos a r r (d) G sin a + H cos a . r r r r The resultant of forces P and Q is R . If Q r is doubled then R is doubled. If the direction r r of Q is reversed,then R is again doubled.

æ 1 1ö 2sçç + ÷÷ è f rø 2s 1 1 + f r

æ1

1ö

(b) 2 sçç + ÷÷ è f rø (d)

2s( f + r )

Two stones are projected from the top of a cliff h metres high , with the same speed u, so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected horizontally and the other is projected at an angle q to the horizontal then tan q equals [2003]

Vector Algebra 2

22.

2u gh

(a) u gh

(b)

u (c) 2 g h

u (d) 2 h g

Two particles start simultaneously from the same point and move along two straight lines, r one with uniform velocity u and the other from r rest with uniform acceleration f . Let a be the

26.

from a point iˆ + 2 ˆj + 3kˆ to the point

27.

(b)

f

(c) 23.

f cos a u

The upper

[2004]

Let u , v , w be such that | u |= 1,| v |= 2,

| u - v + w | equals

29.

Let R1 and R2 respectively be the maximum ranges up and down an inclined plane and R be the maximum range on the horizontal plane.

[2004]

(a) 14

(b)

(c)

(d) 2

14

7

Let a , b and c be non-zero vectors such that (a ´ b ) ´ c =

1 | b || c | a . If q is the acute 3

angle between the vectors b and c , then sinq equals [2004]

Then R1, R , R2 are in [ 2003] (a) H.P (b) A.G..P (c) A.P (d) G..P. r r r Let a, b and c be three non-zero vectors such that no two of these are collinear. If the vector r r r r a + 2b is collinear with c and b + 3cr is collinear with ar (l being some non-zero scalar) r then ar + 2b + 6cr equals [2004] r (a) 0 (b) lb r (c) lc (d) lar

(2l - 1)c are non coplanar for no value of l all except one value of l all except two values of l all values of l

| w |= 3. If the projection v along u is equal to that of w along u and v , w are perpendicular to each other then

3 at a point in the 5

horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is [2003] (a) 80 m (b) 20 m (c) 40 m (d) 60 m.

25.

28.

3 th portion of a vertical pole 4

subtends an angle tan -1

24.

and (a) (b) (c) (d)

u sin a f

(d) u sin a

5iˆ + 4 ˆj + kˆ . The work done in standard units by the forces is given by [2004] (a) 15 (b) 30 (c) 25 (d) 40 r r r If a, b , c are non-coplanar vectors and l is a real number, then the vectors a + 2b + 3c , lb + 4c

angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time [2003] (a) u cos a

M-159 A particle is acted upon by constant forces 4iˆ + ˆj - 3kˆ and 3iˆ + ˆj - kˆ which displace it

(a)

2 2 3

(b)

2 3

2 1 (d) 3 3 With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are [2004]

(c)

30.

1 ö 1 ö æ æ 3 ÷ N and ç 2 3÷ N (a) çè 2 + è 2 ø 2 ø

EBD_7764 M-160

(b)

Mathematics

( 2 + 3 ) N and ( 2 - 3 ) N

34.

1 ö 1 ö æ æ 2 ÷ N and ç 2 2÷ N (c) çè 2 + è 2 ø 2 ø (d) 31.

32.

( 2 + 2 ) N and ( 2 - 2 ) N

In a right angle DABC , ÐA = 90° and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a r force F has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then r magnitude of F is [2004] (a) 9 (b) 4 (c) 5 (d) 3 r r Three forces P, Q and R acting along IA, IB

35.

A B C : cos ec : cos ec 2 2 2

(b) sin

A B C : sin : sin 2 2 2

(c) sec

A B C : sec : sec 2 2 2

A B C : cos : cos 2 2 2 A paticle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5km/hr. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively [2004]

(d) cos

33.

13 17 km / h and km / h (a) 9 9

(b)

13 17 km / h and km / h 4 4

(c)

17 13 km / h and km / h 9 9

17 13 km / h and km / h (d) 4 4

(a)

1 ( 6 - 2 )m / s 8

(b)

1 ( 3 - 1) m / s 4

(c)

1 m/s 4

(d)

1 m/s 8

If t1 and t2 are the times of flight of two particles having the same initial velocity u and range R on the horizontal , then t12 + t22 is equal to

and IC, where I is the incentre of a DABC are r r r in equilibrium. Then P : Q : R is [2004] (a) cos ec

1 m / s is resolved into two 4 components along OA and OB making angles 30° and 45° respectively with the given velocity. Then the component along OB is [2004]

A velocity

[2004]

(a) 1

(b) 4u 2 / g 2

(c) u 2 / 2g

(d) u 2 / g

36.

If C is the mid point of AB and P is any point outside AB, then [2005] uuur uuur uuur (a) PA + PB = 2 PC uuur uuur uuur (b) PA + PB = PC uuur uuur uuur (c) PA + PB + 2 PC = 0 uuur uuur uuur (d) PA + PB + PC = 0

37.

For any vector a , the value of ur ur ur (a ´ iˆ)2 + (a ´ ˆj ) 2 + (a ´ kˆ)2 is equal to

ur 2 (a) 3a

38.

ur 2 (b) a

[2005]

ur 2 ur 2 (c) 2a (d) 4a uur uur uur If a , b , c are non coplanar vectors and l is a real number then [2005] uur uur uur uur uur uur 2 uur uur [l (a + b ) l b l c ] = [a b + c b ] for (a) exactly one value of l (b) no value of l (c) exactly three values of l (d) exactly two values of l

Vector Algebra

39.

40.

uur uur Let a = iˆ - kˆ , b = x iˆ + ˆj + (1 – x) kˆ and uur uur uur uur c = y iˆ + x ˆj + (1 + x – y) kˆ . Then [a , b , c ] depends on [2005] (a) only y (b) only x (c) both x and y (d) neither x nor y ur ur ur ABC is a triangle. Forces P , Q , R acting

44.

resultant of the forces acting along AB, BC with magnitudes

(a)

B A C : sin : sin 2 2 2

(c)

B C A : cos : cos 2 2 2 (d) cos A : cos B : cos C A particle is projected from a point O with velocity u at an angle of 60° with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by [2005]

45.

(c) cos

41.

u (a) 3

(c) 42.

43.

2u 3

(a)

2H A-B

(b)

H A+ B

(c)

H 2( A + B )

(d)

H A-B

The resultant R of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to smaller one is: [2005] (a) 2 : 1

(b) 3 :

(c) 3 : 2

(d) 3 : 2 2

2

2

( AB) ( AC )

(b) ( AB )( AC ) AB + AC

2

1 1 + AB AC ( a ´ b ) ´ c = a ´ (b ´ c )

(d)

1 AD

where

a, b

then a and c are

u

3 A and B are two like parallel forces. A couple of moment Hlies in the plane ofAand Band is contained with them. The resultant ofA and B after combining is displaced through a distance [2005]

If

AB 2 + AC 2

and

c

are

any three vectors such that a .b ¹ 0 , b . c ¹ 0

u (b) 2

(d)

1 1 and respectively AB AC

is the force along AD , where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is [2006]

along IA, IB, and IC respectively are in equilibrium, where I is the incentre of D ABC. Then P : Q : R is [2005] (a) sin A : sin B : sin C (b) sin

M-161 ABC is a triangle, right angled at A. The

46.

[2006]

(a) inclined at an angle of

p between them 3

(b) inclined at an angle of

p between them 6

(c) perpendicular (d) parallel The values of a, for which points A, B, C with position vectors aiˆ - 3 ˆj + kˆ

2iˆ - ˆj + kˆ

,

iˆ - 3 ˆj - 5kˆ

and

respectively are the vertices of a right p are [2006] 2 (b) – 2 and – 1

angled triangle with C = (a) 2 and 1

47.

(c) – 2 and 1 (d) 2 and – 1 A particle has two velocities of equal magnitude inclined to each other at an angle q . If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then q is [2006] (a) 90° (b) 120° (c) 45° (d) 60°

EBD_7764 M-162

48.

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If

54.

g = 10m / s 2 , then the height above the point

49.

P from where the body began to fall is [2006] (a) 720 m (b) 900 m (c) 320 m (d) 680 m If uˆ and vˆ are unit vectors and q is the acute

50.

angle between them, then 2 uˆ ×3 vˆ is a unit vector for [2007] (a) no value of q (b) exactly one value of q (c) exactly two values of q (d) more than two values of q r r Let and a = iˆ + ˆj + kˆ, b = iˆ - ˆj + 2kˆ

51.

52.

r r = c xiˆ + ( x - 2) ˆj - kˆ . If the vector c lies in r r the plane of a and b , then x equals [2007] (a) – 4 (b) – 2 (c) 0 (d) 1. The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be 19 n. The value of P is [2007] (a) 3 n (b) 4 n (c) 5 n (d) 6 n. A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is [2007] (a) tan -1

bc a(c - a)

(b) tan -1

(a) 0

55.

p 4

p (d) p 2 The projections of a vector on the three coordinate axis are 6, –3, 2 respectively. The direction cosines of the vector are : [2009]

(a)

6 -3 2 , , 5 5 5

(b)

6 -3 2 , , 7 7 7

-6 -3 2 , , (d) 6, – 3, 2 7 7 7 r r r If u, v, w are non-coplanar vectors and p, q are real numbers, then the equality r r r r r r rr r [3u pv pw] - [ pv w qu ] - [2w qv qu ] = 0 holds for : [2009] (a) exactly two values of (p, q) (b) more than two but not all values of (p, q) (c) all values of (p, q) (d) exactly one value of (p, q) r r Let a = ˆj - kˆ and c = iˆ - ˆj - kˆ. Then the vector r r r r r r r b satisfying a ´ b + c = 0 and a. b = 3 is[2010]

(c)

56.

57.

(a) 2iˆ - ˆj + 2kˆ

(b) iˆ - ˆj - 2kˆ

(c) iˆ + ˆj - 2kˆ

58.

(d) -iˆ + ˆj - 2kˆ r r If the vectors a = iˆ - ˆj + 2kˆ , b = 2iˆ + 4 ˆj + k%ˆ r and c = liˆ + ˆj + m kˆ are mutually orthogonal, then (l, m) = (a) (2, –3) (c) (3, –2)

(c) tan -1

53.

(b)

(c)

bc a

b (d) 45°. ac A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened to the extremities of a rod 13m long. If the rod be so held that the body hangs immediately below the middle point, then tensions in the strings are [2007] (a) 5 kg and 12 kg (b) 5 kg and 13 kg (c) 12 kg and 13 kg (d) 5 kg and 5 kg

Mathematics r r r The non-zero vectors are a , b and c are r r r r related by a = 8b and c = –7b . Then the angle between ar and cr is [2008]

59.

[2010] (b) (–2, 3) (d) (–3, 2)

r 1 1 r 3iˆ + kˆ and b = 2iˆ + 3 ˆj - 6kˆ , If a = 10 7 r r r é r r r then the value of 2a - b ë a ´ b ´ a + 2b ùû is [2011] (a) –3 (b) 5 (c) 3 (d) –5

(

)

(

(

)(

)

) (

)

Vector Algebra

60.

61.

r r The vectors a and b are not perpendicular r r and c and d are two vectors satisfying r r r r rr b ´ c = b ´ d and a.d = 0 . Then the vector r [2011] d is equal to r rr r æ b .cr ö r r æ a.c ö r b + ç r r ÷c r c + b (a) (b) çr ÷ è a.b ø è a.b ø r rr r æ b.cr ö r r æ a.c ö r b ç r r ÷c (c) c - ç r r ÷ b (d) è a.b ø è a.b ø If

the

pi$ + $j + k$ , $i + q $j + k$

63.

64.

5iˆ – 2 ˆj + 4kˆ are the sides of a triangle ABC, then the length of the median through A is [2013] (a) (b) 18 72 (c)

66.

and

iˆ + ˆj + rkˆ ( p ¹ q ¹ r ¹ 1) vector are coplanar,,

62.

65.

then the value of pqr - ( p + q + r ) is [2011RS] (a) 2 (b) 0 (c) – 1 (d) – 2 r r r Let a, b, c be three non-zero vectors which r r are pairwise non-collinear. If a + 3b is collinear r r r r with c and b + 2c is collinear with a , then r r r a + 3b + 6c is : [2011RS] r r (a) a (b) c r r r (c) 0 (d) a + c r r Let a and b be two unit vectors. If the vectors r r c = aˆ + 2 bˆ and d = 5 aˆ - 4bˆ are perpendicular to each other, then the angle between aˆ and [2012] bˆ is : p p (a) (b) 6 2 p p (c) (d) 3 4 Let ABCD be a parallelogram such that uuur r uuuur r AB = q , AD = p and ÐBAD be an acute angle. r If r is the vector that coincide with the altitude directed from the vertex B to the side AD, then r [2012] r is given by : rr r r r r 3 ( p.q ) r ( p .q ) (a) r = 3q - r r p (b) rr = -qr + r r pr ( p.p ) ( p . p) r r r r r r 3 ( p.q ) r r r (p.q) r (c) r = q - r r p (d) r = -3q - r r p (p.p) (p .p)

M-163 uuur uuur If the vectors AB = 3iˆ + 4kˆ and AC =

67.

(d) 33 45 r r r r r r rrr 2 If éë a ´ b b ´ c c ´ a ùû = l éë a b c ùû then l is equal to [2014] (a) 0 (b) 1 (c) 2 (d) 3 ® ®

®

Let a , b and c be three non-zero vectors such that no two of them are collinear and ®

®

®

(a ´ b) ´ c =

1® ®® b c a . If q is the angle 3 ®

®

between vectors b and c , then a value of sin q is : [2015]

68.

(a)

2 3

(c)

2 2 3 ® ®

®

Let a , b and c be three unit vectors such that ® æ® ® ö ® 3 æç® ® ö a ´ çç b´ c ÷÷ = b+ c ÷ . If b is not parallel ç ÷ è ø 2 è ø ®

®

69.

-2 3 3 - 2 (d) 3 (b)

®

to c , then the angle between a and b is: [2016] 2p 5p (a) (b) 3 6 3p p (c) (d) 4 2 r r r ˆ ˆ ˆ ˆ Let a = 2i + j - 2k and b = i + ˆj . Let c be a r r r r r vector such that | c - a | = 3, a ´ b ´ c = 3 r r and the angle between rc and a ´ b be 30°.

(

rr

Then a.c is equal to : 1 25 (a) (b) 8 8 (c) 2 (d) 5

)

[2017]

EBD_7764 M-164

Mathematics Answer Key

1 (b) 16 (none ) 31 (c) 46 (a) 61 (d)

1.

2 (a) 17 (c) 32 (d) 47 (b) 62 (c)

3 (a) 18 (c) 33 (d) 48 (a) 63 (c)

4 (a) 19 (c) 34 (a) 49 (b) 64 (b)

5 (a) 20 (a) 35 (b) 50 (b) 65 (c)

6 (b) 21 (a) 36 (a) 51 (d) 66 (b)

7 (c) 22 (a) 37 (c) 52 (c) 67 (c)

8 (a) 23 (c) 38 (b) 53 (a) 68 (b)

3 =4 3. 2

® ®

4.

® ®

12 (d) 27 (c) 42 (b) 57 (d)

® ® ® (a) We have, ® a+ b+ c = 0

®

® ®

®

®

® ® ® ® ® ®

+ 2( a × b + b × c + c × a ) = 0

® ® ® ® ®

(a) We have, [ a ´ b b x c

® ® ® ® ® ®

c ´ a]

Þ 25 + 16 + 9 + 2( a × b + b × c + c × a ) = 0

® ® ì ® ® ® ®ü = ( a ´ b ). í( b x c ) ´ ( c ´ a )ý î þ

® ® ® ® ® ®

Þ ( a × b + b × c + c × a ) = -25 .

® ® ì ® ® ® ® ® ®ü = ( a ´ b ). í( m × a ) c - ( m × c ) a )ý î þ

® ® ®® ®®

5.

® ®

(where m = b ´ c ) ® ® ®

\ | a × b + b × c + c × a |= 25 . r r r (a) Since a , c , b form a right handed system, iˆ ˆj kˆ r r r \ c = b ´ a = 0 1 0 = ziˆ - xkˆ

= {( a ´ b ). c }.{( a × ( b ´ c )}

x

®®® = [ a b c ]2 = 42 = 16 .

3.

(a)

®

®

® ®

®

6.

®

a+ b+ c = 0 Þ b+ c = - a ® ®

®

®

® ®

®®

Þ 2 | b | | c | cos q = 49 - 34 = 15 ;

y ®

z ®

(b) We have a ´ b = 39 k = c ®

Þ ( b + c ) 2 = ( a ) 2 = 52 + 32 + 2 b × c = 7 2 ®

15 (c) 30 (c) 45 (d) 60 (c)

Þ | a |2 + | b |2 | + | c |2

Þ ( a ´ b )2 = 16

® ® ®

14 (b) 29 (a) 44 (d) 59 (d)

® ® ®

® ®

®

13 (c) 28 (c) 43 (d) 58 (d)

Þ ( a + b + c )2 = 0

Þ ( a ´ b ) 2 + 48 = 16 ´ 4

®

11 (d) 26 (d) 41 (d) 56 (d)

p Þ cos q = 1/2; Þ q = = 60° 3

Now, ( a ´ b )2 + ( a × b )2 = a 2b 2 ;

2.

10 (a) 25 (c) 40 (c) 55 (b)

Þ 2 × 5 × 3cos q = 15;

p ® ® ® (b) We have, a . b = | a | | ® b | cos 6

= 4 ´ 2´

9 (b) 24 (a) 39 (d) 54 (a) 69 (c)

®

®

Also | a |= 34,| b |= 45, | c |= 39 ; ®

®

®

\ | a |:| b |:| c |= 34 : 45 : 39 .

M-165

Vector Algebra

7.

8.

r r r r (c) Let a + b + c = r . Then r r r r r r a ´ (a + b + c ) = a ´ r r r r r r r Þ0+a´b +a´c =a ´r r r r r r r r r r Þ a ´b -c ´a = a ´r Þ a ´r =0 r r r r r r Similarly b ´ r = 0 & c ´ r = 0 Above three conditions will be satisfied r r for non-zero vectors if and only if r = 0 (a) Given P + Q = 18 .......(1) P2 + Q2 + 2PQ cos a = 144 ......(2)

tan 90º =

Q sin a P + Q cos a

nˆ =

j

k

1

1

0

1 -1 0

=

-2 kˆ = - kˆ 2

2´ 2 r w.nˆ = (i + 2 j + 3k ).( - k ) = -3 = 3

11.

(d)

r r r F + F1 + F2 = 7i + 2 j - 4k r r r d = PV . of B - PV . of A = 4i + 2 j - 2k r r W = F .d = 28 + 4 + 8 = 40 unit

12. (d)

A

3i + 4 k

R=12

Q

i

5i – 2j + 4k

a B

P

uuur (3 + 5)i + (0 - 2) j + (4 + 4) k P.V of AD = 2 uuur = 4i - j + 4k or AD = 16 + 16 + 1 = 33

Þ P + Q cos a = 0 ......(3) From (2) and (3), Q2 – P2 = 144 Þ (Q – P) (Q + P) = 144 144 =8 18 From (1), On solving, we get Q = 13, P = 5 (b) Ð TQW = 180 – q ; Ð RQW = 2q ; Ð RQT = 180 – q P

\

9.

Q-P=

q T O

q

Q 90–2q R

w Applying Lami's theorem at Q.

10.

13. (c)

14.

q

T R W = = sin 2q sin(180 - q) sin(180 - q) Þ R = W and T = 2W cos q r r r (a) sin ce n is perpendicular u and v , r r r u´n n= r r | u || n |

C

D

r r r r r r r r r a + b + c = 0 Þ (a + b + c ).( a + b + c ) = 0 r2 r2 r2 rr rr rr a + b + c + 2(a.b + b .c + c .a ) = 0

r r r r r r -1 - 4 - 9 = -7 a.b + b .c + c .a = 2 (b) Vector perpendicular to the face OAB

iˆ ˆj kˆ uuur uuur = OA ´ OB = 1 2 1 = 5iˆ - ˆj - 3kˆ 2 1 3

Vector perpendicular to the face ABC

ˆj iˆ uuur uuur = AB ´ AC = 1 -1 -2 -1

kˆ 2 = iˆ - 5 ˆj - 3kˆ 1

Angle between the faces = angle between their normals cos q =

5+5+9 35 35

=

19 æ 19 ö or q = cos -1 ç ÷ 35 è 35 ø

EBD_7764 M-166

18. (c)

a a 2 1 + a3

15.

(c)

c2

1 + c3

a a2 1 Þb b

2

c

2

c

r G = rp sin q

r H = rp cos q éQ sin(90o + q) = cos qù ë û

b b 2 1 + b3 = 0 c

Mathematics

r r r G=r ´ p;

G = rp sin q.......(1)

a a2

a3

1+ b b

2

3

b =0

1

2

c3

c

c

H = rp cos q........(2)

x = rp sin( q + a)...... ( 3 )

From (1), (2) & (3), r r x = G cos a + H sin a . 19. (c) R 2 = P 2 + Q 2 + 2 PQ cos q

1 a a2 Þ (1 + abc) 1 b b 2 = 0 1 c

c

2

1 a a2 As 1 b b 2 ¹ 0 ( given condition ) 1 c

c2

\ abc = -1

16.

(none) A = (7,-4,7), B = (1,-6,10), C = (-1, - 3, 4) and D = (5, –1, 5)

AB = (7 - 1)2 + ( -4 + 6)2 + (7 - 10)2 = 36 + 4 + 9 = 7

17.

Similarly BC = 7, CD = 41, DA = 17 \ None of the options is satisfied r r r r r r r r r r r (c) (u + v - w).(u ´ v - u ´ w - v ´ v + v ´ w) r r r r r r r r r = (u + v - w).(u ´ v - u ´ w + v ´ w) r r r = u.(u ´ v ) r r r r r r r r r r r r -u.(u ´ w) + u.(v ´ w) + v .(u ´ v ) - v .(u ´ w) r r r r r r r r r r r r +v .(v ´ w) - w.(u ´ v ) + w.(u ´ w) - w.(v ´ w) r r r r r r r r r = u.(v ´ w) - v .(u ´ w) - w.(u ´ v ) rrr rr r r rr r r r = [uvw)] + [vwu )] - [ wuv ] = u.(v ´ w)

2

2

.......(1)

2

4 R = P + 4Q + 4 PQ cos q

.......(2)

4 R 2 = P 2 + Q 2 - 2 PQ cos q 2

2

On (1) + (3), 5R = 2 P + 2Q

........(3) 2

........(4)

On (3) ´ 2 + (2), 12 R 2 = 3 P 2 + 6Q 2 .....(5) 2 P 2 + 2Q 2 - 5 R 2 = 0

.....(6)

3P 2 + 6Q 2 - 12R 2 = 0 2

2

.....(7) 2

P Q R = = -24 + 30 24 - 15 12 - 6

.

P 2 Q2 R 2 = = or P 2 : Q 2 : R 2 = 2 : 3 : 2 6 9 6 20. (a) Let the body travels from A to B with constant acceleration t and from B to C with constant retardation r. y x t1 t2 B A C If AB = x, BC = y, time taken from A to B = t1 and time taken from B to C = t2, then s = x + y and t = t1 + t2 For the motion from A to B v 2 = u 2 + 2 fs Þ v 2 = 2 fx (Q u = 0 )

v2 2f and v = u + ft Þ v= ft1 v Þ t1 = f For the motion from B to C

Þ

x=

....(1)

...(2)

v 2 = u 2 + 2 fs

Þ

0 =v2 – 2ry Þ y =

v2 2r

...(3)

M-167

Vector Algebra

and v = u + ft Þ 0 = v – rt2 v Þ t2 = r Adding equations (1) and (3), we get

1 and h = -u sin q ´ t + gt 2 2 From (1) and (2) we get

v2 é 1 1 ù x+ y = ê + ú = s 2 ë f rû Adding equations (2) and (4), we get é 1 1ù t1 + t 2 = v ê + ú = t ë f rû

1 cos q

h=-

æ 1 1ö t = 2s ç + ÷ è f rø (a) For the stone projected horizontally, for horizontal motion, using distance = speed × time Þ R = ut and for vertical motion u 1 h = 0 ´ t + gt 2 2

u sin q 2h 1 é 2h ù + gê ú cos q g 2 êë g cos 2 q ûú

h = -u

2h tan q + h sec 2 q g

h = -u

2h tan q + h tan 2 q + h g

Þ

21.

Þ t =h

\

2h g

Substituting this value of t in eq (3) we get

2

é 1 1ù v2 ê + ú 2 t ë f rû = 1 + 1 = 2s v2 æ 1 1 ö f r 2´ ç + ÷ 2 è f rø

\

2h = u cos q ´ t g

u

Þ t=

....(3)

2h g

2 2 tan q = 0; \ tan q = u hg hg

tan 2 q - u

22. (a) We can consider the two velocities as r r $ $ v1 = ui$ and v2 = ( ft cos a ) i + ( ft sin a ) j

ft

2h ....(1) We get RR= u g

a

For the stone projected at an angle q, for horizontal and vertical motions, we have

u

\ Relative velocity of second with respect to first r r r v = v2 - v1 = ( ft cos a - u ) $i + ft sin a $j

u q

Þ h

r2 2 2 v = ( ft cos a - u ) + ( ft sin a )

2 2 2 = f t + u - 2uft cos a r For v to be min we should have 2

R R = u cos q ´ t

....(2)

Þ

dv = 0 Þ 2 f 2t - 2uf cos a = 0 dt u cos a t= f

EBD_7764 M-168

Mathematics

2

Also \ v

d v

2

dt 2 2

r r r r r r r Þ a + 2b + 6c = tc + sa - b + 3c

r r r r r r = tc + (b + 3c ) - b + 3c = (t + 6)c

= 2 f 2 = + ve

and hence v is least at the time

u cos a f 23.

­

(c)

3 h 4

¯ ­

b a

h 4

q 40 m

tan q - tan a 1 + tan q.tan a

tan b =

or

h h 3 40 160 = h h 5 1+ . 40 160

0 h2 - 200h + 6400 = 40 or 160 metre Þ h= \ possible height = 40 metre

24.

(a) Let b be the inclination of the plane to the horizontal and u be the velocity of projection of the projectile R1 =

u2 u2 and R2 = g (1 + sin b) g (1 - sin b)

1 1 2g 1 1 2é u2 ù + = 2 or + = êQ R = ú R1 R2 u R1 R2 R êë g úû \ R1 , R, R2 arein H .P.

25.

r r r r r r (c) Let a + 2b = tc and b + 3c = s a, where t and s are scalars. Adding, we get r r r r r a + 3b + 3c = tc + sa

r = lc , where l = t + 6 r 26. (d) Resultant of forces F = 7iˆ + 2 ˆj - 4kˆ r Displacement d = 4iˆ + 2 ˆj - 2kˆ r r \ Work done = F .d = 28 + 4 + 8 = 40 r r r r r 27. (c) Vectors a + 2b + 3c , lb + 4c , and r (2l - 1)c are coplanar if

¯

æ3ö q = a + b , b = tan -1 ç ÷ è5ø or b = q - a

Þ

r r r é using s a = b + 3c ù ë û

1 2

3

0 l

4

=0

0 0 2l - 1 1 2 \ Forces are noncoplanar for all l, except

Þ l(2l - 1) = 0 Þ l = 0 or

l = 0,

1 2

rr rr r r v . u v .u 28. (c) Projection of v along u = r = |u | 2 r r rr r r w. u w.u projection of w along u = r = |u | 2 rr rr v .u w.u = Given ....(1) 2 2 Also , vr.wr = 0 ....(2) r r r Now | u - v + w |2 r 2 r2 r 2 rr r r rr = | u | + | v | + | w | -2u.v - 2v .w + 2u.w = 1 + 4 + 9 + 0 [ From (1) and (2)] = 14 r r r \| u - v + w |= 14 r r r 1 r r r 29. (a) Given (a ´ b ) ´ c = | b || c | a 3 r r Clearly a and b are noncollinear

rr r rr r 1 r r r Þ (a.c )b - (b .c )a = | b | | c | a 3

M-169

Vector Algebra

rr 1 r r rr \ a.c = 0 and - b.c = | b || c | 3 -1 Þ cos q = 3

\ sin q = 1 -

30.

P

1 2 2 = 9 3

I

r r [q is acute angle between b and c ] (c) Let forces be P and Q. then P + Q = 4 ....(1) and P 2 + Q 2 = 32 Solving we get the forces

31.

A

....(2)

æ æ 2ö 2ö ç 2 + 2 ÷ N and ç 2 - 2 ÷ N è ø è ø (c) Since, the moment about A is zero, hence r F passes through A. Taking A as origin. r Let the line of action of force F be y = mx . (see figure)

Moment about B = Y

3m 1+ m

2

90+

A 2

R

B

C

P A sin(90° + ) 2

=

Q Bö æ sin ç 90° + ÷ 2ø è

=

R Cö æ sin ç 90° + ÷ 2ø è

A B C : cos : cos 2 2 2 33. (d) Time taken by the particle in complete journey P : Q : R = cos

Þ

T=

r | F |= 9 ....(1)

12 5 + = 4 hr. 4 5

5 km

C(0,4)

F (y = mx)

B(3,0) r 4 | F |= 16....(2) Moment about C= 1 + m2 A

Dividing (1) by (2), we get

32.

Q

v 3 Þ| F |= 5 N . 4 (d) IA, IB, IC are bisectors of the angles A, B and C as I is incentre of DABC.

A

B

12km

\ Average speed =

12 + 5 17 = 4 4

12 2 + 5 2 13 = 4 4 [using vector addition]

Average velocity =

1 , component along OB 4

m=

34. (a) If v =

B C A - = 90° + etc. 2 2 2 Applying Lami’s theorem at I

1 1 ´ v sin 30° 6- 2 = = 4 2 = sin(45° + 30°) 8 3 +1 2 2 35. (b) For same horizontal range the angles of

Now ÐBIC = 180 -

projection must be a and

\ t1 =

2u sin a and g

p -a 2

EBD_7764 M-170

Mathematics

l ( a1 + b1 ) l ( a2 + b2 ) l ( a3 + b3 )

æp ö 2u sin ç - a ÷ è2 ø 2u cos a t2 = = g g

\ t12 + t22 =

36.

4u

Þ

l 2b1

l 2b2

lb3

lc1

lc2

lc3

2

g2

a1

uuur uuur uuur uuur (a) PA + AP = 0 and PC + CP = 0 uuuur uuuur uuuur Þ PA + AC + CP = 0 uuuur uuuur uuuur and PB + BC + CP = 0 Adding, we get uuuur uuuur uuuur uuuur uuur PA + PB + AC + BC + 2CP = 0. uuuur uuuur uuuur uuuur Since AC = - BC & CP = - PC uuuur uuuur uuuur Þ PA + PB - 2PC = 0.

a2

b1

Þl

4

37.

38.

b1

b2

b3

c1

c2

c3

a2

)

a3

= b1 + c1 b2 + c2 b1

B

r c = c1i$ + c2 $j + c3kˆ then as per question r r r r é l ar + b l 2 b lcr ù = é ar b + cr b ù û ë û ë

(

a3 + b3

b3 + c3

b2

b3

R1 - R2

r r r r (c) Let a = xi + yj + zk r r r r r r a ´ i = zj - yk Þ (a ´ i )2 = y 2 + z 2 Similarly, r r r r (a ´ j )2 = x 2 + z 2 and (a ´ k )2 = x 2 + y 2 r r r r r r Þ (a ´ i )2 + (a ´ j )2 + (a ´ k )2 r = 2( x 2 + y 2 + z 2 ) = 2a 2 (b) Let us consider r a = a1$i + a2 $j + a3kˆ r b = b1$i + b2 $j + b3kˆ

b3

a1 + b1 a2b2

a1 a2

C

b3 + c3

b2

a1

P

A

a3

= b1 + c1 b 2 + c2

39.

R2 - R3 a3

a1 a2

a3

Þ l 4 b1

b2

b3 = c1

c2

c3

c1

c2

c3

b2

b3

b1

Þ l 4 = -1 Hence l has no real values. r r (d) a = iˆ - kˆ, b = xiˆ + ˆj + (1 - x)kˆ and r c = yiˆ + xjˆ + (1 + x - y )kˆ

1 0 r rr rr r [a b c ] = a.b ´ c = x 1 y

-1 1- x

x 1+ x - y

= 1 é1 + x - y - x + x 2 ù - é - x 2 - y ù ë û ë û = 1 - y + x 2 - x2 + y =1 rrr Hence éë a b c ùû is independent of x and y both. 40. (c) IA, IB, IC are bisectors of the angles A, B and C as I is incentre of DABC. B C A Now ÐBIC = 180 - - = 90° + etc. 2 2 2 Applying Lami’s theorem at I

M-171

Vector Algebra

and AC respectively are

A

æ 1 ö ˆ æ 1 öˆ ÷ j çè ÷ i and çè AC ø AB ø \

P I Q

90+

æ 1 ö æ 1 ö =ç i+ j è AB ÷ø çè AC ÷ø

R

A 2

\

B Sin(90° +

A ) 2

=

Q Bö æ sin ç 90° + ÷ è 2ø

Þ P : Q : R = cos

(d)

Magnitude of resultant is

C P

41.

Their resultant along AD

2

R

=

æ 1 ö æ 1 ö = ç +ç è AB ÷ø è AC ÷ø

Cö æ sin ç 90° + ÷ è 2ø

A B C : cos : cos 2 2 2

=

u cos 60° = v cos 30° (as horizontal component of velocity remains the same) 1 3 1 Þ u× = v× or v = u 2 2 3

2

BC AB. AC

1 AC

^j

D q

u 60° O

42.

30° o

o

ucos60

30

o

vcos30 X

BC AC BC 1 = = Þ AB AD AB ´ AC AD \ The required magnitude of resultant

Þ

H A+ B F ' = 3F cos q and F = 3F sin q

becomes 45. (d)

F

B

But from figure DABC ~ DDBA

(b) Let A and B be displaced by a distance x then Change in moment of (A + B) =applied moments

(d)

1 ^ i AB

v

Þ ( A + B) ´ x = H Þ x = 43.

q

A

o

30

o

AB 2 + AC 2

C

Y

60

AC 2 + AB 2

=

1 . AD

( a ´ b ) ´ c = a ´ ( b ´ c ) , a .b ¹ 0 , b .c ¹ 0

3F

Þ (a . c ).b - ( b . c )a = (a . c ).b - (a . b ).c Þ (a . b ).c = ( b . c ) a Þ a || c .

F'

Þ F ' = 2 2 F Þ F : F ' :: 3 : 2 2 . 44.

(d) If we consider unit vectors iˆ and ˆj in the direction AB and AC respectively, then as per quesiton, forces along AB

46. (a)

uuur CA = (2 - a )iˆ + 2 ˆj ; uuur CB = (1 - a)iˆ - 6 kˆ uuur uuur CA.CB = 0

Þ (2 - a)(1 - a) = 0 a = 2, 1 Þ

EBD_7764 M-172

47.

(b) For two velocities u and u at an angle q to each other the resultant is given by D

C

u

h

R E

q/4 R¢

q/2

q/2 q/4 u

A

48.

Mathematics 1 1 2 (a) Using h = gt 2 and h + 400 = g (t + 4) 2 2

R2 = u2 + u2

400m

B

2u2 cosq = 2u2

(1 + cos q) q Þ R 2 = 4u 2 cos 2 q / 2 or R = 2u cos 2 Now in second case, the new resultant AE (i.e., R¢) bisects ÐCAB , therefore using angle bisector theorem in DABC , we get AB BE u u/2 = Þ = ÞR=u AC EC R u/2 q Þ 2u cos = u 2 q 1 q Þ cos = = cos 60° Þ = 60° 2 2 2 or q = 120° +

Q(t)

u/2

u sin q q tan = 2 4 u + u cos q 2

q 1 q 1 q sin + sin cos q = sin q cos 4 2 4 2 4 q 3q q q 2 sin = sin = 3sin - 4sin 3 4 4 4 4 1 q q sin 2 = Þ = 30° or q = 120° 4 4 4 R2 R1

P(t+4) Subtracting, we get 400 = 8g + 4gt Þ t = 8 sec 1 \ h = ´ 10 ´ 64 = 320m 2 \ Desired height = 320 + 400 = 720 m 49. (b) Given | 2uˆ ´ 3vˆ | = 1 and q is acute angle between uˆ and vˆ , | uˆ | = 1, | vˆ | = 1 Þ 6 | uˆ | | vˆ | | sin q | = 1

1 6 Hence, there is exactly one value of q for which 2 uˆ × 3 vˆ is a unit vector.. r r 50. (b) Given a = iˆ + ˆj + kˆ , b = $i - $j + 2k$ and r c = xi$ + ( x - 2) $j - k$ r r r If c lies in the plane of a and b , then rrr [a b c ] = 0

1

Þ \

\

u

qq 44

q/ 2 u/2

u

6 | sin q | = 1 Þ sin q =

Þ

1

1

-1 2 =0 i.e. 1 x ( x - 2) -1 Þ 1[1 – 2(x – 2)] – 1[– 1 – 2x] + 1[x – 2 + x] = 0 Þ 1 – 2x + 4 + 1 + 2x + 2x – 2 = 0 Þ 2x = –4 Þ x = – 2

51. (d)

8r r Clearly a = – c 7 r r Þ a || c and are opposite in direction r r \ Angle between a and c is p. r r rr a.c 8b × (–7b ) cos q = r r = r r = –1 Þ q = p ac 8b 7b

M-173

Vector Algebra 52. (c) Given : Force P = Pn, Q = 3n, resultant R =

é aù = a tan a ê1 - ú ë Rû æ u 2 sin 2 a ö R Q = ç ÷ g è ø

7n & P' = Pn, Q' = (–3)n, R' = 19 n

Ö19

P

é aù b = a tan a ê1 - ú ë cû æ c - aö Þ b = a tan a . ç è c ÷ø bc Þ tan a = a(c - a) The angle of projection, bc a = tan–1 a(c - a)

7

Þ

a 3

–3

We know that R2 = P2 + Q2 + 2PQ cos a Þ (7)2 = P2 + (3)2 + 2 × P × 3 cos a Þ 49 = P2 + 9 + 6P cos a Þ 40 = P2 + 6P cos a .....(i) 2 2 2 19 = P + (–3) + 2P × –3 cos a and Þ Þ

53.

( )

19 = P2 + 9 – 6P cos a 10 = P2 – 6P cos a .....(ii) Adding (i) and (ii) 50 = 2P2 Þ P2 = 25 Þ P = 5n. (a) Let B be the top of the wall whose coordinates will be (a, b). Range (R) = c u

A

A

12m

a

D

M B q 9

C c

ga é ù b = a tan a ê1 ú 2 2 ë 2u cos a tan a û a é ù = a tan a ê1 - 2 ú ê 2u cos 2 a. sin a ú g êë cos a úû

a é ù 1- 2 ê ú = a tan a u .2sin a cos a ê ú g ëê ûú

a é ù 1= a tan a ê u 2 sin 2a ú ê ú g êë úû

q 0–

T1 C 5m 13 kg

b

a

13 m

q

B (a,b)

B lies on the trajectory x2 1 y = x tan a – g 2 2 u cos 2 a a2 1 Þ b = a tan a – g 2 2 u cos 2 a Þ

54. (a)

Q 132 = 5 2 + 122 Þ AB2 = AC2 + BC2 Þ ÐACB = 90° Q m is mid point of the hypotenuse AB, therefore MA = MB = MC Þ ÐA = ÐACM = q Applying Lami’s theorem at C, we get

13kg T1 T2 = = sin(180 - q) sin(90 + q) sin 90° Þ T1 = 13 sin q and T2 = 13 cos q 5 12 and T2 = 13 ´ 13 13 Þ T1 = 5 kg and T2 = 12 kg 55. (b) Let P (x1, y1, z1) and Q (x2, y2, z2) be the initial and final points of the vector whose projections on the three coordinate axes are 6, – 3, 2 then x2 – x1, = 6 ; y2 – y1 = – 3 ; z2 – z1 = 2 uuur So that direction ratios of PQ are 6, – 3, 2

Þ

T1 = 13 ´

EBD_7764 M-174

\

uuur Direction cosines of PQ are 6 2

2

2

2

2

6 + ( - 3) + 2

-3

,

2

6 + ( - 3)2 + 22

2 2

56.

=

,

6 –3 2 , , 7 7 7

6 + ( - 3) + 2 r r r (d) Q u , v , w are non coplanar vectors r r r \ [ u , v , w] ¹ 0 Now, r r r r r r r r r [ 3u , pv , p w]–[ pv , p w, qu ] [ 2 w, qv , qu ] = 0 r r r r r r Þ 3 p2 [ u , v , w] – pq [ v , w, u ] – 2q2 r r r [ w, v , u ] = 0 r r r r r r r r r Þ 3 p2 [ u , v , w] - pq [ u , v , w] + 2q 2 [ u , v , w]

r r r

Þ (3 p2 – pq + 2q2) [ u , v , w ] = 0 Þ 3p2 – pq + 2q2 = 0 Þ 2p2 + p2 – pq +

q 2 7 q2 + =0 4 4

2

qö 7 2 æ Þ çè p - ÷ø + q = 0 2 4 Þ p = 0, q = 0, p = q / 2 This is possible only when p = 0, q = 0 \ There is exactly one value of (p, q). rr r r r rr r (d) cr = b ´ ar Þ b. c = b.(b ´ a ) Þ b . c = 0 2p2 +

57.

(

)(

Mathematics r r 58. (d) Since, a, b and cr are mutually orthogonal r r r r r r \ a . b = 0, b . c = 0 , c . a = 0

Þ 2l + 4 + m = 0

l - 1 + 2m = 0

(

)

Þ b2 - b3 = 3 From equation (i) b1 = b2 + b3 = (3 + b3 ) + b3 = 3 + 2b3 r b = (3 + 2b3 )iˆ + (3 + b3 ) ˆj + b3 kˆ From the option given, it is clear that b3 equal to either 2 or –2. r If b3 = 2 then b = 7iˆ + 5 ˆj + 2kˆ which is not possible r If b = -2, then b = -iˆ + ˆj - 2kˆ 3

...(ii)

On solving (i) and (ii), we get l = -3, m = 2 r r r r r r 59. (d) (2a - b). (a ´ b ) ´ (a + 2b ) r r r r r r r r = (2 a - b ). ( a ´ b ) ´ a + 2( a + b ) ´ b r r r r r r r r r r r æ (a . a )b - (a . b )a + 2(a . b )b ö = (2a - b ) ç r r r÷ è -2(b . b ) a ø r r r r = (2 a - b )(b - 0 + 0 - 2 a ) r r r r [Q a . b = 0 and b . b = 1] rr rr = -4 a.a - b .b = -5 r rr 60. (c) a.b ¹ 0 , ar.d = 0 r r r r Now, b ´ c = b ´ d r r r r r r Þ a ´ (b ´ c ) = a ´ (b ´ d ) r r r r r r r r r r r r Þ ( a . c )b - ( a . b ) c = ( a . d ) b - ( a . b ) d rr r rr r rr r Þ (a.b )d = -(a.c )b + (a.b )c r r æ ar.cr ö r d = c - ç r r ÷b è a.b ø 61. (d) The given vectors are collinear if

Þ b1iˆ + b2 ˆj + b3 kˆ . iˆ - ˆj - kˆ = 0, r where b = b1iˆ + b2 ˆj + b3 kˆ b1 - b2 - b3 = 0 ...(i) r r and a. b = 3 Þ ( ˆj - kˆ).(b1iˆ + b2 ˆj + b3kˆ) = 3

...(i)

)

(

)

p 1 1 1 q 1 =0 1 1 r Þ Þ Þ

62. (c)

p ( qr - 1) + 1(1 - r ) + 1(1 - q) = 0 pqr - p + 1 - r + 1 – q = 0

pqr - ( p + q + r ) = -2 r r r a + 3b = l c

.............(i) r r r ............(ii) b + 2c = μa On solving equation (i) and (ii) r r (1 + 3μ ) a - (λ + 6) c = 0 r r As a and c are non collinear,, \ 1 + 3μ = 0 and l + 6 = 0 r r r r From (i), a + 3b + 6c = 0

Vector Algebra r r 63. (c) Let c = aˆ + 2 bˆ and d = 5 aˆ - 4bˆ r r Since c and d are perpendicular to each other r r \ c .d = 0

( aˆ + 2bˆ ) .( 5aˆ - 4bˆ ) = 0

Þ

64.

r r Þ 5 + 6 aˆ.bˆ - 8 = 0 ( Q a .a = 1) p 1 Þ aˆ.bˆ = Þ q = 3 2 (b) Let ABCD be a parallelogram such that uuur r uuuur r AB = q , AD = p and ÐBADbe an acute angle. We have B

65.

Þ Þ Þ

Q

q

A

C

67. (c)

r X

p

D

uuur æ pr . qr öæ pr ö pr.qr r AX = ç r ÷ç r ÷ = r 2 p ç p ÷ç p ÷ è øè ø p rr uuur uuur uuur r p.q r r Let r = BX = BA + AX = -q + r p 2 p (c) We have, uuur uuur uuur uuur uuur uuur AB + BC + CA = 0 Þ BC = AC - AB uuur uuur uuur uuuur AC - AB æ uuuur BC ö Now, BM = çQ BM = 2 ÷ 2 è ø

Þ 68. (b)

M-175 r r rrr r r rr = (a ´ b).[[b c a ] c] [Q b ´ c. c = 0] r rr r r r rrr 2 = [a b c].(a ´ b. c ) = [a b c ] r r r r r r rrr = [a ´ b b ´ c c ´ a ] = [a b c ]2 So l = 1 r r r 1 r rr a´b ´c = b c a 3 r r r 1 r rr –c ´ a ´ b = b c a 3 rr r rr r 1 r r r – c.b a + c.a b = b c a 3 r r r rr r 1 r rr – b c cos qa + c.a b = b c a 3 r r r a, b, c are non collinear, the above equation is possible only when rr 1 – cosq = and c.a = 0 3 1 2 2 cosq = – Þ sinq = ; qÎ II quad 3 3 r r r r r 3 a ´ (b ´ c) = (b + c) 2 r r r r r r 3r 3r Þ (a × c)b - (a × b)c = b+ c 2 2

(

)

(

) ( ) ( )

( )

On comparing both sides r r 3 3 a ×b = Þ cosq= 2 2 r r [Q a and b are unit vectors] r r where q is the angle between a and b 5p 6 Given : r r ˆ b = ˆi + ˆj a = 2iˆ + ˆj - 2k, r Þ | a |= 3 r r \ a ´ b = 2iˆ - 2jˆ + kˆ r r | a ´ b |= 22 + 22 + 12 = 3 r r r r r r We have (a ´ b) ´ c = | a ´ b || c | sin 30 n r 1 r r r r 1 Þ | (a ´ b) ´ c | = 3| c |. Þ 3 = 3 | c | . 2 2 r \ |c| =2 r r Now | c - a | = 3 On squaring, we r rget rr Þ c2 + a2 – 2 – c.a = 9 Þ 4 + 9 – 2 – a.c = 9 rr rr rr Þ a.c = 2 [Q c.a = a.c ]

q=

69. (c)

Also, we have uuur uuuur uuuur AB + BM + MA = 0 uuur uuur uuur AC - AB uuuur = AM Þ AB + 2 uuur uuur uuuur AB + AC = 4iˆ - ˆj + 4kˆ Þ 1 AM = 2 uuuur AM = 33 Þ 66.

r r r r r r (b) L.H.S = (a ´ b).[(b ´ c) ´ (c ´ a)] r r r r r r r rr r = (a ´ b).[(b ´ c. a )c - (b ´ c.c )a ]

EBD_7764 M-176

Mathematics

26

Three Dimensional Geometry 1.

A plane which passes through the point (3, 2,

6.

x 2 + y 2 + z 2 + 2 x - 2 y - 4 z - 19 = 0 is cut by the

x-4 y -7 z -4 is[2002] = = 1 5 4 (a) x – y + z =1 (b) x + y + z = 5

0) and the line

2.

(c) x + 2y – z = 1 (d) 2x – y + z = 5 The d.r. of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle p /4 with plane x + y = 3 are [2002] (a) 1, 2 ,1

3.

(b) 1, 1,

The radius of the circle in which the sphere

7.

plane x + 2 y + 2 z + 7 = 0 is [2003] (a) 4 (b) 1 (c) 2 (d) 3 Two system of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a' , b' , c ' from the origin then (a)

2

(c) 1, 1, 2 (d) 2 , 1, 1 The shortest distance from the plane to the sphere 12 x + 4 y + 3z = 327

(b) (c)

x 2 + y 2 + z 2 + 4 x - 2 y - 6 z= 155 is [2003]

(a) 39

(b) 26

(d)

4 (d) 13. 13 The two lines x = ay + b , z = cy + d and x = a¢y + b¢, z = c¢y + d¢ will be perpendicular, if and only if [2003] (a) aa¢ + cc¢ + 1 = 0 (b) aa¢ + bb¢ + cc¢ + 1 = 0 (c) aa¢ + bb¢ +cc¢ = 0 (d) (a + a¢) (b + b¢) +(c + c¢) = 0.

(c) 11

4.

5.

The lines x - 2 = y - 3 = z - 4 1

and

1

-k

[2003]

x -1 y - 4 z - 5 = = are coplanar if k 1 1

(a) k = 3 or –2 (c) k = 1 or –1

(b) k = 0 or –1 (d) k = 0 or –3.

8.

1 a

2

1 a2 1 a

2

1 a2

+

+ + -

1 b

2

1 b2 1 b

2

1 b2

+

+ -

1 c

2

1 c2 1 c

2

1 c2

-

+ + +

1 a'

2

1 a'2 1 a'

2

1 a'2

-

1 b'

1

+ +

2

b' 2 1 b'

-

2

1 b'2

-

+ -

1 c' 2

1 c '2 1 c'2 1 c'2

[2003] =0

=0 =0 =0.

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y +4z + 5 = 0 is [2004] (a)

9 2

(b)

5 2

7 3 (d) 2 2 A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and

(c) 9.

x + a = 2 y = 2 z . The co-ordinates of each of the points of intersection are given by [2004] (a) (2a,3a, 3a), (2a, a, a ) (b) (3a, 2a,3a), (a, a, a) (c) (3a, 2a, 3a), (a, a, 2a ) (d) (3a,3a,3a), (a, a, a)

Three Dimensional Geometry 10. If the straight lines

[2004]

15.

x = 1 + s , y = -3 - l s , z = 1 + l s t , y = 1 + t , z = 2 - t , with parameters 2 s and t respectively, are co-planar, then l equals. (a) 0 (b) –1

x 2 + y 2 + z 2 + 6 x - 8 y - 2 z = 13 and

and x =

11.

1 (c) (d) –2 2 The intersection of the spheres

x 2 + y 2 + z 2 - 10 x + 4 y - 2 z = 8 then a equals

16.

x 2 + y 2 + z 2 + 7 x - 2 y - z = 13 and x 2 + y 2 + z 2 - 3x + 3 y + 4 z = 8 is the same as

the intersection of one of the sphere and the plane [2004] (a) 2 x - y - z = 1 (b) x - 2 y - z = 1 12.

(c) x - y - 2 z = 1 (d) x - y - z = 1 A line makes the same angle q, with each of the x and z axis. If the angle b, which it makes with

(a) (c) 13.

(b)

3 5

(d)

17.

18.

(a)

(b)

-3 5

Let a, b and c be distinct non- negative

ciˆ + cjˆ + bkˆ lie in a plane, then c is

[2005]

3 -4 (d) 4 3 The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is [2005] (a) 0° (b) 90° (c) 45° (d) 30°

(c) 14.

5 3

æ 1ö (d) ç1,- ÷ 2ø è

numbers. If the vectors aiˆ + ajˆ + ckˆ , iˆ + kˆ and

x +1 y -1 = 1 2

1 then the value of l is 3

3 3

x y 1 + + = 0 always passes a b c through a fixed point. That point is [2005] (a) (– 1, 2) (b) (– 1, – 2)

z -2 = and the plane 2x – y + l z + 4 = 0 is such 2

that sin q =

10

10 3 (d) 3 10 If non zero numbers a, b, c are in H.P., then the

(c) (1, – 2)

2 3

(b)

straight line

1 5

If the angle q between the line

10 9

(c)

[2004]

2 5

[2005] (a) – 1 (b) 1 (c) – 2 (d) 2 The distance between the line ur r = 2iˆ - 2 ˆj + 3kˆ + l(i - j + 4k ) and the plane ur r .(iˆ + 5 ˆj + kˆ) = 5 is [2005] (a)

y-axis, is such that sin 2 b = 3 sin 2 q, then cos2q equals

M-177 If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres

19.

[2005]

(a) the Geometric Mean of a and b (b) the Arithmetic Mean of a and b (c) equal to zero (d) the Harmonic Mean of a and b The plane x + 2y – z = 4 cuts the sphere x 2 + y 2 + z 2 – x + z – 2 = 0 in a circle of radius

[2005] (a) 3

(b) 1

(c) 2

(d)

2

EBD_7764 M-178

20.

The two lines x = ay + b , z = cy + d ; and x = a ' y + b ' , z = c ' y + d ' are perpendicular to each other if (a) aa '+ cc ' = -1

[2006] (b) aa '+ cc ' = 1

26.

a c (d) a + c = 1 + = -1 a' c' a' c' The image of the point (–1, 3, 4) in the plane

(c) 21.

x - 2 y = 0 is

22.

(b) (15,11, 4)

17 19 (c) æç - , - ,1ö÷ 3 ø è 3

(d) None of these

If a line makes an angle of p / 4 with the positive directions of each of x- axis and y- axis, then the angle that the line makes with the positive direction of the z-axis is [2007] (a)

p 4

æ 17 –13ö çè 0, , ÷ . Then 2 2 ø (a) a = 2, b = 8 (c) a = 6, b = 4

[2006]

æ 17 19 ö (a) ç - , - , 4 ÷ 3 ø è 3

(b)

p 2

p p (d) 3 6 If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the cooordinates of the other end of the diameter are [2007] (a) (4, 3, 5) (b) (4, 3, – 3) (c) (4, 9, – 3) (d) (4, –3, 3). Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle a with the positive x-axis, then cos a equals [2007]

27.

24.

(a) 1 (c) 25.

(b)

1 3

(d)

28.

29.

1 2 1 . 2

r The vector a = a iˆ + 2 ˆj + bkˆ lies in the plane r r of the vectors b = iˆ + ˆj and c = ˆj + kˆ and r bisects the angle between b and cr . Then

If the straight lines

[2008] (b) a = 4, b = 6 (d) a = 8, b = 2 x –1 y – 2 z – 3 = = and k 2 3

x – 2 y – 3 z –1 = = intersect at a point, then 3 k 2 the integer k is equal to [2008] (a) –5 (b) 5 (c) 2 (d) –2

(c)

23.

Mathematics which one of the following gives possible values of a and b? [2008] (a) a = 2, b = 2 (b) a = 1, b = 2 (c) a = 2, b = 1 (d) a = 1, b = 1 The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz-plane at the point

30.

x – 2 y –1 z + 2 = = lie in the 3 –5 2 plane x + 3y – az + b = 0. Then (a, b) equals [2009] (a) (–6, 7) (b) (5, –15) (c) (–5, 5) (d) (6, –17) Statement -1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. Statement -2: The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). [2010] (a) Statement -1 is true, Statement -2 is true ; Statement -2 is not a correct explanation for Statement -1. (b) Statement -1 is true, Statement -2 is false. (c) Statement -1 is false, Statement -2 is true . (d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1. A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle q with the positive z-axis, then q equals [2010] (a) 45° (b) 60° (c) 75° (d) 30°

Let the line

M-179

Three Dimensional Geometry

31.

x y + = 1 passes through 5 b the point (13, 32). The line K is parallel to L and

35.

x y + = 1 . Then the distance c 3 between L and K is [2010]

(c) 32.

17

23 17

(b)

(d)

(a)

15 23

36.

15 y -1 z - 3 = 2 l

æ 5 ö and the plane x + 2y + 3z = 4 is cos–1 ç ÷, è 14 ø

then l equals (a)

33.

3 2

37.

2 5

5 2 (c) (d) 3 3 Statement-1: The point A(1, 0, 7)) is the mirror image of the point B(1, 6, 3) in the line :

33

(b)

2 9

9 (d) 0 2 Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is [2013] (c)

38.

(a)

3 2

(b)

5 2

(c)

7 2

(d)

9 2

Statement-2: The line

34.

(b)

(c) (d) 53 66 A equation of a plane parallel to the plane x – 2y + 2z –5 = 0 and at a unit distance from the origin is : [2012] (a) x – 2y + 2z – 3 = 0 (b) x – 2y + 2z + 1 = 0 (c) x – 2y + 2z – 1 = 0 (d) x – 2y + 2z + 5 = 0 x -1 y +1 z-1 = = and If the line 2 3 4

(a) –1

x y -1 z - 2 = = 1 2 3

x y -1 z - 2 = = 1 2 3 bisects the line segment joining A(1, 0, 7) and B(1, 6, 3) . [2011] (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. The distance of the point (1, – 5, 9) from the plane x – y + z = 5 measured along a straight x = y = z is [2011RS]

29

[2011RS]

x-3 y-k z = = intersect, then k is equal to: 1 2 1 [2012]

[2011] (b)

(3, -1, 11) to the line

x y - 2 z -3 = = is : 2 3 4

17

If the angle between the line x =

(c) 3 10 (d) 3 5 The length of the perpendicular drawn from the point

has the equation

(a)

(b) 5 3

(a) 10 3

The line L given by

39.

If the lines

x - 2 y -3 z - 4 = = and 1 1 -k

x -1 y -4 z -5 = = are coplanar, then k k 2 1 can have [2013] (a) any value (b) exactly one value (c) exactly two values (d) exactly three values

EBD_7764 M-180

40.

The image of the line

x -1 y - 3 z - 4 in = = 3 1 -5

44.

the plane 2 x - y + z + 3 = 0 is the line: [2014] (a)

x-3 y +5 z - 2 = = 3 1 -5

(b)

x-3 y +5 z - 2 = = -3 -1 5

(c)

x+3 y -5 z - 2 = = 3 1 -5

45.

l 2 + m 2 + n 2 is

46.

(b)

p 2

p p (d) 3 4 The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is : [2015] (a) x + 3y + 6z = 7 (b) 2x + 6y + 12z = – 13 (c) 2x + 6y + 12z = 13 (d) x + 3y + 6z = –7 The distance of the point (1, 0, 2) from the point

43.

47.

x -3 y+2 z+4 = = lies in the -1 2 3 2 2 plane, lx + my – z = 9, then l + m is equal to : [2016] (a) 5 (b) 2 (c) 26 (d) 18 If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to

If the line,

(c) 2 42 (d) 42 The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines

x -1 y + 2 z - 4 and = = -2 1 3

x - 2 y +1 z + 7 = = , is : 2 -1 -1

(a)

x - 2 y +1 z - 2 = = of intersection of the line 3 4 12 and the plane x – y + z = 16, is [2015] (a) 3 21 (b) 13

(c) 2 14

(d) 10 3

x y z = = is Q, then PQ is equal to : 1 4 5 (a) 6 5 (b) 3 5

(c) 42.

20 3

line,

[2014]

p 6

(b)

3

(c) 3 10

x+3 y -5 z + 2 = = -3 -1 5 The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and

(a)

10

(a)

(d) 41.

Mathematics The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is : [2016]

(c)

10

(b)

74 10

(d)

83

[2017]

20 74 5 83

(d) 8

Answer Key 1 (a) 16 (b) 31 (c) 46 (c)

2 (b) 17 (c) 32 (d) 47 (c)

3 (d ) 18 (a) 33 (a)

4 (a) 19 (b) 34 (a)

5 (d) 20 (a) 35 (c)

6 (d) 21 (d) 36 (a)

7 (a) 22 (b) 37 (c)

8 (c) 23 (c) 38 (c)

9 (b) 24 (c) 39 (c)

10 (d) 25 (d) 40 (c)

11 (a) 26 (c) 41 (c)

12 (c) 27 (a) 42 (a)

13 (a) 28 (a) 43 (b)

14 (b) 29 (a) 44 (d)

15 (c) 30 (b) 45 (b)

M-181

Three Dimensional Geometry

1.

2.

(a) As the point (3, 2, 0) lies on the given line x-4 y -7 z -4 = = 1 5 4 \ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points (3, 2, 0) and (4, 7, 4) \ x – y + z = 1 is the required plane. (b) Equation of plane through (1, 0, 0) is a (x – 1) + by + cz = 0 ...(i) (i) passes through (0, 1, 0). –a + b = 0 Þ b = a; Also,

cos 45° =

a+a 2

2(2a + c )

Þ 2a2 = c2 3.

2

Þ c=

2a .

-2 ´ 12 + 4 ´ 1 + 3 ´ 3 - 327 144 + 9 + 16

(d) O

centre of sphere = (-1, 1, 2) Radius of sphere 1 + 1 + 4 + 19 = 5 Perpendicular distance from centre to the plane -1 + 2 + 4 + 7 12 = = 4. OC = d = 3 1+ 4 + 4

7.

- 4 + 1 + 9 + 155

AC 2 = AO 2 - OC 2 = 52 - 4 2 = 9 Þ AC = 3 x y z (a) Eq. of planes be + + =1 & a b c x y z + + =1 a' b' c ' ( ^ r distance on plane from origin is same.)

-1 = 1 1 + + a 2 b 2 c2 1

= 26 – 13 = 13 4.

5.

(a) x - b = y = z - d ; x - b ' = y = z - d ' . a 1 c a' 1 c' For perpendicularity of lines aa'+1 + cc' = 0 x2 - x1

y2 - y1

z2 - z1

l1

m1

n1

l2

m2

n2

(d)

1 - 1 -1 1

1

-k = 0 Þ

k

2

1

C

A

Þ 2a = 2a2 + c2

So d.r of normal are a, a 2a i.e. 1, 1, 2 . (d) Shortest distance = perpendicular distance between the plane and sphere = distance of plane from centre of sphere – radius =

6.

1

8.

0

2

1+ k

k+2

1

2

+

1

2

-

or 2 x + y + 2 z +

-1 -k = 0 1

k 2 + 3k = 0 Þ k (k + 3) = 0 or k = 0 or - 3

a'

1

2

-

1

+

2

-

2

b'

+

9.

1 c '2

1

5 =0 2 \ Distance between (1) and (2) 5 +8 2 21 7 = = = 2 2 2 2 +1 + 2 2 9 2

=0

0

1

2

-1 1

= 0 a b c a' b' c' 2 (c) The planes are 2 x + y + 2 z - 8 = 0 . ...(1) and 4 x + 2 y + 4 z + 5 = 0 2

+

1

...(2)

(b) Let a point on the line x = y + a = z is (l, l - a, l ) and a point on the line

m mö æ x + a = 2 y = 2 z is ç m - a, , ÷ , then è 2 2ø

EBD_7764 M-182

Mathematics

direction ratio of the line joining these m m points are l - m + a, l - a - , l 2 2 If it respresents the required line, then l-m+a = 2

m m l2 = 2 1 2

2

Þ 2 cos q = 3 - 3 cos q

13.

l-a-

(

2a 2a ö æ (3a, 3a-a,3a) and çè 2a - a, , ÷ø 2 2

and 2 x = y - 1 =

Þ sin q =

..........(1)

z -2 =t -1

The lines are coplanar, if

.........(2)

0 - ( -1) -1 - 3 -2 - ( -1) 1 -l l =0 1 1 -1 2 1 c2 ® c2 + c3 ; 1 1 2

11.

12.

-5 -1 0 l =0 0

-1

l Þ 5( -1 - ) = 0 Þ l = -2 2 (a) The equations of spheres are

)(

14.

2 l

=

1 Þ 4l = 5 + l 3

3 5 +l 5 Þl= . 3 (b) The given lines are 2 x = 3 y = - z x y z = = 3 2 -6 and 6 x = - y = -4 z

or

[Dividing by 6]

x y z = = [Dividing by 12] 2 -12 -3 \ Angle between two lines is

or

3.2 + 2.( -12) + ( -6) . ( -3)

cos q =

32 + 22 + ( -6) =

6 - 24 + 18 49 157

2

2 2 + ( -12 ) + ( -3) 2

2

= 0 Þ q = 90o

15. (c) Plane 2ax - 3ay + 4az + 6 = 0 passes through the mid point of the centre of spheres

S1 : x 2 + y 2 + z 2 + 7 x - 2 y - z - 13 = 0 and

x 2 + y 2 + z 2 + 6 x - 8 y - 2 z = 13 and

S2 : x 2 + y 2 + z 2 - 3x + 3 y + 4 z - 8 = 0 Their plane of intersection is S1 - S2 = 0 Þ 10 x - 5 y - 5 z - 5 = 0

x 2 + y 2 + z 2 - 10 x + 4 y - 2 z = 8 respectively centre of spheres are (– 3, 4, 1) and (5, – 2, 1). Mid point of centres is (1, 1, 1). Satisfying this in the equation of plane, we get 2a - 3a + 4a + 6 = 0 Þ a = -2.

Þ 2x - y - z =1 (c) The direction cosines of the line are cos q, cos b, cos q \ cos 2 q + cos 2 b + cos 2 q = 1 Þ 2 cos 2 q = sin 2 b = 3 sin 2 q (given)

)

ö 2-2+2 l æp cos ç - q ÷ = 2 ø 3´ 5 + l è

or (3a, 2a,3a) and (a,a,a) (d) The given lines are y + 3 z -1 x -1 = = =s -l l

3 5 (a) If q is the angle between line and plane æp ö then çè - q÷ø is the angle between line 2 and normal to plane given by $i + 2 $j + 2kˆ . 2$i - $j + l kˆ æp ö cos ç - q÷ = è2 ø 3 4 +1+ l \ cos 2 q =

on solving we get l = 3a, m = 2a \ The required points of intersection are

10.

2

M-183

Three Dimensional Geometry 16. (b) The given line is r r r r r r r r = 2i - 2 j + 3k + l (i - j + 4k ) r r r r and the plane is r × (i + 5 j + k ) = 5 or x + 5y + z = 5 Required distance

=

17.

2 - 10 - 2 + 3 - 5 1 + 25 + 1

(c) a, b, c are in H.P. Þ Þ

=

20. (a) Equation of lines x - b = y = z - d a 1 c

10 3 3

1 1 1 , , are in A.P.. a b c

2 1 1 1 2 1 = + Þ - + =0 b a c a b c

x y 1 + + = 0 passes through (1, –2) a a c r r r r r (a) Vector ai + aj + ck , i + k and r r r ci + cj + bk are coplanar

a a c 1 0 1 = 0 Þ c 2 = ab Þ c = ab c c b

19.

(b)

\ c is G.M. of a and b.

3 2

5 2

1 , 0, – 1 2 2

Perpendicular distance of centre 1ö æ1 ç ,0,- ÷ 2 2ø è from x + 2y – 2 = 4 is given by

1 1 + -4 = 2 2

3 2

\ radius of circle =

a -1 æ b + 3 ö - 2ç ÷=0 2 è 2 ø

Also line joining (a , b, g ) and ( –1, 3, 4) should be parallel to the normal of the plane x –2y = 0 a +1 b - 3 g - 4 \ = = =l -2 1 0 Þ a = l - 1, b = -2l + 3, g = 4 … (2) From (1) and (2) 9 13 a= , b=- , g =4 5 5 None of the option matches. 22. (b) Let the angle of line makes with the positive direction of z-axis is a direction cosines of line with the +ve directions of x-axis, y-axis, and z-axis is l, m, n respectively. p p \ l = cos , m = cos , n = cos a 4 4 as we know that, l2 + m2 + n2 = 1 p p \ cos2 + cos2 + cos2 a = 1 4 4 1 1 Þ + + cos2 a = 1 2 2 p Þ cos2 a = 0 Þ a = 2 Hence, angle with positive direction of the p . 2 23. (c) We know that equation of sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where centre is (–u, –v, –w) given x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0 \ centre º (3, 6, 1) Coordinates of one end of diameter of the sphere are (2, 3, 5). Let the coordinates of

z-axis is

6

radius of sphere =

\

\ a - 1 - 2b - 6 = 0 Þ a - 2b = 7 … (1)

\

18.

x - b' y z - d ' = = a' 1 c' Line are perpendicular Þ aa '+ 1 + cc ' = 0 21. (d) If ( a, b, g ) be the image, then mid point of (a, b, g) and (–1, 3, 4) must lie on x – 2y = 0

1 1 + +2 = 4 4 5 3 - =1. 2 2

5 2

EBD_7764 M-184

the other end of diameter are (a, b, g )

24.

\ a + 2 = 3, b+3 = 6 , g +5 =1 2 2 2 Þ a = 4, b = 9 and g = –3 \ Coordinate of other end of diameter are (4, 9, –3) (c) Let the direction cosines of line L be l, m, n, then 2l + 3m + n = 0 ....(i) and l + 3m + 2n = 0 ....(ii) on solving equation (i) and (ii), we get l m n = = 6 - 3 1- 4 6 - 3

Now

l m n = = = 3 -3 3

Þ

l m n = = 3 -3 3

l 2 + m 2 + n2 32 + ( -3) 2 + 32

Q l2 + m2 + n2 = 1 l m n 1 \ = = = 3 -3 3 27

3 1 1 1 = ,m = ,n = 27 3 3 3 Line L, makes an angle a with +ve x-axis Þ l=

\ l = cos a Þ cos a =

25.

1 3

r r r (d) Q a lies in the plane of b and c r r r \ a = b + lc Þ aiˆ + 2 ˆj + bkˆ = iˆ + ˆj + l ( ˆj + kˆ) Þ a = 1, 2 = 1+ l, b = l Þ a = 1, b = 1 r r r Q a bisects the angle between b and c . r \ a = l(bˆ + cˆ) ˆ l (iˆ + 2ˆj + k) Þ aˆi + 2ˆj + bkˆ = 2

Þa=

26.

l

, l = 2 , b=

l

2 2 Þ a=b=1 (c) Equation of line through (5, 1, a) and x – 5 y –1 z – a = = =l –2 b –1 1– a \ Any point on this line is a [–2l + 5, (b – 1) l + 1, (1– a) l + a]

(3, b, 1) is

Mathematics It crosses yz plane where –2l + 5 = 0

l=

5 2

5 5 ö æ 17 –13ö æ \ çè 0,(b –1) + 1, (1– a) + a÷ø = çè 0, , ÷ø 2 2 2 2

5 17 5 13 Þ (b – 1) + 1 = and (1 – a ) + a = 2 2 2 2 Þ b = 4 and a = 6 27. (a) The two lines intersect if shortest distance between them is zero i.e. r r r r (a2 – a1 ) × b1 ´ b2 =0 r r b1 ´ b2 r r r r Þ (a2 – a1) × b1 ´ b2 = 0 r where a1 = iˆ + 2 ˆj + 3kˆ , r b1 = kiˆ + 2 ˆj + 3kˆ r a2 = 2iˆ + 3 ˆj + kˆ , bˆ2 = 3iˆ + kjˆ + 2kˆ

1 1 –2

Þ k 2 3 k

3 =0 2

Þ 1(4 –3k) –1(2k – 9) – 2(k2– 6) = 0 Þ –2k2 – 5k + 25 = 0 Þ k = –5 or Q k is an integer, therefore k = –5 28. (a) Q The line

5 2

x - 2 y -1 z + 2 = = lie in the -5 3 2

plane x + 3y – a z + b = 0 \ Pt (2, 1, – 2) lies on the plane i.e. 2 + 3 + 2a + b = 0 or 2a + b + 5 = 0 ....(i) Also normal to plane will be perpendicular to line, \ 3 × 1 – 5 × 3 + 2 × (– a ) = 0 Þ a=–6 From equation (i) then, b = 7 \ (a, b) = (– 6, 7) 29. (a) A(3, 1, 6); B = (1, 3, 4) Mid-point of AB = (2, 2, 5) lies on the plane. and d.r’s of AB = (2, –2, 2) d.r’s of normal to plane = (1, –1, 1). Direction ratio of AB and normal to the plane are proportional therefore, AB is perpendicular to the plane

M-185

Three Dimensional Geometry

30.

31.

32.

\ A is image of B Statement-2 is correct but it is not correct explanation. (b) Direction cosines of the line : -1 , 1 , m = cos120° = l = cos 45° = 2 2 n = cos q where q is the angle, which line makes with positive z-axis. Now l 2 + m2 + n2 = 1 1 1 2 Þ + + cos q = 1 2 4 1 cos 2 q = 4 1 cos q = ( q being acute) Þ 2 p Þ q= 3 b (c) Slope of line L = 5 3 Slope of line K = c Line L is parallel to line k. b 3 Þ = Þ bc = 15 5 c (13, 32) is a point on L. 13 32 32 8 + =1 Þ =\ 5 b b 5 3 Þ b = -20 Þ c = 4 Equation of K : y - 4 x = 3 Þ 4x - y + 3 = 0 Distance between L and K 52 - 32 + 3 23 = = 17 17 (d) If q be the angle between the given line and plane, then 1´1+ 2 ´ 2 + l ´ 3 sin q = 12 + 22 + l 2 . 12 + 22 + 32 5 + 3l = 14. 5 + l 2 Þ cos q = 1 -

(5 + 3l ) 2 14(5 + l 2 )

2 \ q = cos -1 1 - (5 + 3l ) 14(5 + l 2 )

But it is given that q = cos -1

\ 1-

33.

34.

(5 + 3l ) 2 2

=

5 14

5 14

14(5 + l 2 Þ l= 3 (a) The direction ratio of the line segment joining points A(1, 0, 7) and B(1, 6, 3) is 0, 6, –4. The direction ratio of the given line is 1, 2, 3. Clearly 1 × 0 + 2 × 6 + 3 × (–4) = 0 So, the given line is perpendicular to line AB. Also , the mid point of A and B is (1, 3, 5) which lies on the given line. So, the image of B in the given line is A, because the given line is the perpendicular bisector of line segment joining points A and B. (a) Equation of line through P (1, -5,9) and parallel to the plane x = y = z is x -1 y + 5 z - 9 = = = l ( say ) 1 1 1 Q = ( x = 1 + l, y = -5 + l, z = 9 + l ) Given plane x – y + z =5 \ 1+ l + 5 - l + 9 + l = 5 Þ l = -10 \ Q = (– 9 , – 15, – 1)

\

PQ =

(1 + 9) 2 + (15 - 5) 2 + ( 9 + 1) 2

= 300 = 10 3 35. (c) Let feet of perpendicular is

( 2a, 3a + 2, 4a + 3) Þ Direction ratio of the ^ line is 2a - 3,3a + 3, 4a - 8. and Direction ratio of the line 2, 3, 4 are Þ 2 ( 2α –3) +3 (3α + 3) +4 (4α – 8) =0 Þ 29a - 29 = 0 Þ a =1 Þ Feet of ^ is (2, 5, 7)

36.

Þ Length ^ is 12 + 6 2 + 42 = 53 (a) Given equation of a plane is x – 2y + 2z – 5 = 0 So, Equation of parallel plane is given by x – 2y + 2z + d = 0 Now, it is given that distance from origin to the parallel plane is 1. d \ =1 Þ d = ±3 2 1 + 2 2 + 22

EBD_7764 M-186

Mathematics

So equation of required plane x – 2y + 2z ± 3 = 0 x -1 y +1 z -1 = = 37. (c) Given lines are 2 3 4 x-3 y-k z = = and 1 2 1 r r r r Thus, a , b , c and d are given as r r r a ( 1, -1,1 ) , b ( 2, 3, 4 ) , c ( 3, k , 0 ) ; and r d ( 1, 2,1) These lines will intersect if lines are coplanar r r r r i.e., a - c , b & d are coplanar r r r r \ ëé a - c , b , d ûù = 0 r r Now, a - c = (3 – 1, k + 1, 0 –1) = (2, k + 1, –1) 2 k + 1 -1 Þ 2

38.

1 2 1 Þ 2 (3 – 8) – k + 1 (2 – 4) – 1 (4 – 3) = 0 Þ 2 (–5) – ( k +1) (–2) – 1 (1) = 0 Þ – 10 + 2k + 2 – 1 = 0 Þ k = 9 2 (c) 2x + y + 2z – 8 = 0 …(Plane 1) 5 =0 …(Plane 2) 2 Distance between Plane 1 and 2

2x + y + 2z +

= 39.

5 2

22 + 12 + 22

=

-21 7 = 6 2

(c) Given lines will be coplanar -1 1 1 If 1 1 - k = 0 k 2

Þ –1(1 + 2k) – (1 + k2) + 1(2 – k) = 0 Þ k = 0, –3 40.

(c)

3i + j - 5k P 3i + j - 5k A (a, b, c) æ a +1 b + 3 c + 4 ö , , P =ç ÷ 2 2 ø è 2 6-l l +8ö æ , = ç l + 1, ÷ 2 2 ø è 6-l l +8 + +3= 0 2 2 3l + 6 = 0 Þ l = –2 a = –3, b = 5, c = 2

\

4 =0

3

-8 -

A (1, 3, 4)

a -1 b - 3 c - 4 = = = l (let) -1 2 1 Þ a = 2l + 1 b=3–l c=4+l

2(l + 1) -

x+3 y -5 z - 2 = = 3 1 -5 (c) Given, l + m + n = 0 and l 2 = m2 + n2 Now, (–m –n)2 = m2 + n2 Þ mn = 0 Þ m = 0 or n = 0 If m = 0 then l = –n

Required line is

41.

We know l 2 + m2 + n2 = 1 Þ n = ±

1 2

1 ö æ 1 i.e. (l1, m1, n1) = ç , 0, ÷ è 2 2ø If n = 0 then l = –m l 2 + m2 + n2 = 1 Þ 2m2 = 1

1

Þ m=± Let m =

2

1 2

Þl=-

1 2

and n = 0

æ 1 1 ö , , 0÷ (l2, m2, n2 ) = ç è 2 2 ø

p 1 Þ q= 3 2 42. (a) Equation of the plane containing the lines 2x – 5y + z = 3 and x + y + 4z = 5 is 2x – 5y + z – 3 + l (x + y + 4z – 5) = 0 Þ (2 + l) x + (–5 + l) y+ (1 + 4l)z + (–3 – 5l) = 0 ...(i) Since the plane (i) parallel to the given plane x

\

cos q =

M-187

Three Dimensional Geometry

+ 3y + 6z = 1 2 + l -5 + l 1 + 4l \ = = 1 3 6 11 Þ l=2 Hence equation of the required plane is

11ö 55 ö æ 11ö æ æ 44 ö æ çè 2 - ÷ø x + çè -5 - ÷ø y + çè1 - ÷ø z + çè -3 + ÷ø 2 2 2 2 =0 Þ (4 - 11)x + (-10 - 11)y + (2 - 44)z + ( -6 + 55) = 0

Þ distance OP = 10 3 45. (b) Line lies in the plane Þ (3, –2, –4) lie in the plane Þ 3l – 2m + 4 = 9 or 3l – 2m = 5 ..... (1) Also, l, m,–1 are dr's of line perpendicular to plane and 2, –1, 3 are dr's of line lying in the plane Þ 2l – m – 3 = 0 or 2l – m = 3 .....(2) Solving (1) and (2) we get l = 1 and m = –1 Þ l2 + m2 = 2. 46. (c) Equation of line PQ is x -1 y + 2 z - 3 = = 1 4 5

Þ -7x - 21y - 42z + 49 = 0 Þ x + 3y + 6z - 7 = 0 43.

Let F be (l + 1, 4l - 2, 5l + 3)

Þ x + 3y + 6z = 7 (b) General point on given line º P(3r + 2, 4r – 1, 12r + 2) Point P must satisfy equation of plane (3r + 2) – (4r – 1) + (12r + 2) = 16 11r + 5 = 16 r=1 P(3 × 1 + 2, 4 × 1 – 1, 12 × 1 + 2) = P(5, 3, 14) distance between P and (1, 0, 2)

P(1, –2, 3)

F

D = (5 - 1)2 + 32 + (14 - 2)2 = 13 44.

(d)

Q P(1, –5, 9)

Since F lies on the plane \ 2 (l + 1) + 3 (4l – 2) – 4 (5l + 3) + 22 = 0 2l + 2 + 12l – 6 – 20l – 12 + 22 = 0 Þ –6l + 6 = 0 Þ l = 1 \ F is (2, 2, 8)

x=y=z

47. (c)

Q O

x -1 y + 5 z - 9 = = =l 1 1 1 Þ x =l+ 1; y =l- 5; z =l+ 9. Putting these in eqn of plane :l+ 1 -l+ 5 +l+ 9 = 5 Þl= -10 Þ O is (–9, –15, –1) eqn of PO :

PQ = 2 PF = 2 12 + 42 + 52 = 2 42 Let the plane be a (x – 1) + b (y + 1) + c (z + 1) = 0 Normal vector ˆi ˆj kˆ 1 -2 3 = 5iˆ + 7ˆj + 3kˆ 2 -1 -1 So plane is 5 (x – 1) + 7 (y + 1) + 3 (z + 1) = 0 Þ 5x + 7y + 3z + 5 = 0 Distance of point (1, 3, –7) from the plane is 5 + 21 - 21 + 5 10 = 25 + 49 + 9 83

EBD_7764 M-188

Mathematics

27

Probability 1.

A problem in mathematics is given to three students A, B, C and their respective probability 1 1 1 , and . 2 3 4 Probability that the problem is solved is [2002] 3 1 (a) (b) 4 2 2 1 (c) (d) 3 3 A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002] (a) 8/3 (b) 3/8 (c) 4/5 (d) 5/4 The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P (X = 1) is [2003] 1 1 (a) (b) 32 4 1 1 (d) (c) 16 8 4 The probability that A speaks truth is , while 5 3 the probability for B is . The probability that 4 they contradict each other when asked to speak on a fact is [2004] 4 1 (a) (b) 5 5 7 3 (c) (d) 20 20 A random variable X has the probability distribution:

of solving the problem is

2.

3.

4.

5.

X: 1 p(X): 0.2

2 3 4 5 0.2 0.1 0.1 0.2

6 7 8 0.1 0.1 0.1

6.

7.

8.

9.

For the events E = {X is a prime number } and F = { X < 4}, the P( E È F ) is [2004] (a) 0.50 (b) 0.77 (c) 0.35 (d) 0.87 The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is [2004] 28 219 (a) (b) 256 256 128 37 (d) (c) 256 256 Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005] 2 1 (a) (b) 9 9 8 7 (c) (d) 9 9 A random variable X has Poisson distribution with mean 2. Then P (X > 1.5) equals [2005] 2 (a) 2 (b) 0 e 3 3 (c) 1 (d) 2 e e2 At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is [2006] 6 5 (a) (b) e 6 5 (c)

6 55

(d)

6 e5

Probability 10. Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is [2007] (a) 0.2 (b) 0.7 (c) 0.06 (d) 0.14. 11. A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007] (a) 8/729 (b) 8/243 (c) 1/729 (d) 8/9. 12. It is given that the events A and B are such that

1 1 2 P ( A) = , P ( A | B ) = an d P ( B | A) = . 4 2 3 Then P(B) is [2008]

13.

(a)

1 6

(b)

1 3

(c)

2 3

(d)

1 2

(a)

9 , then n is greater than: 10

16.

31 , then p lies in the interval 32

æ 3 11 ù (a) çè , ú 4 12 û

17.

15.

[2011]

é 1ù (b) ê0, ú ë 2û

æ 11 ù æ 1 3ù (c) çè ,1ú (d) çè , ú 12 û 2 4û If C and D are two events such that C Ì D and P(D) ¹ 0, then the correct statement among the following is [2011] (a) P (C | D ) ³ P (C )

(c) P (C | D) =

18.

(d) P (C | D ) = P (C ) Let A, B, C, be pairwise independent events

(

)

P Ac Ç B c / C .

1 9 (b) log 4 – log 3 log10 4 + log10 3 10 10

4 1 (d) log10 4 – log10 3 log10 4 – log10 3 One ticket is selected at random from 50 tickets numbered 00,01,02,...,49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals: [2009] 1 5 (a) (b) 7 14 1 1 (c) (d) 50 14 An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is [2010]

P( D ) P (C )

with P (C) > 0 and P ( A Ç B Ç C ) =0. Then

[2009]

[2011RS]

( ) P( A ) - P(B )

c (a) P B – P ( B )

(c) 14.

2 1 (b) 7 21 2 1 (c) (d) 23 3 Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least one failure is greater than or equal to

(a)

(b) P (C | D ) < P (C )

1ö æ In a binomial distribution B ç n, p = ÷ , if the è 4ø probability of at least one success is greater than or equal to

M-189

(c) 19.

c

c

c

( )

c (d) P A - P ( B )

Three numbers are chosen at random without replacement from {1,2,3,..8}. The probability that their minimum is 3, given that their maximum is 6, is : [2012] 3 1 (a) (b) 8 5 2 1 (d) 5 4 A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is: [2013]

(c)

20.

c

( ) ( )

(b) P A + P B

EBD_7764 M-190

(a) (c) 21.

17

13

(b)

5

3

11

35 10

(d)

5

22.

3 35 Let A and B be two events such that

(

)

P AÈ B =

(

)

1 1 , P A Ç B = and 4 6

23.

( )

1 P A = , where A stands for the 4 complement of the event A. Then the events A and B are [2014] (a) independent but not equally likely. (b) independent and equally likely. (c) mutually exclusive and independent. (d) equally likely but not independent.

Mathematics A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, oneby-one, with replacement, then the variance of the number of green balls drawn is : [2014]

(a)

6 25

(b)

12 5 (c) 6 (d) 4 If two different numbers are taken from the set (0, 1, 2, 3, ......., 10), then the probability that their sum as well as absolute difference are both multiple of 4, is : [2014]

(a)

7 55

(b)

6 55

(c)

12 55

(d)

14 55

Answer Key 1 (a) 16 (b)

1.

2 (d) 17 (a)

(a) P(E1) =

3 (b) 18 (d)

4 (c) 19 (b)

5 (b) 20 (c)

6 (a) 21 (a)

7 (b) 22 (b)

1 1 1 , P (E2) = and P (E3) = ; 3 2 4 _

_

8 (c) 23 (b)

4.

_

P ( E1UE2UE3 ) = 1 - P ( E1 ) P ( E 2 ) P ( E 3 )

2.

3.

æ 1ö æ 1ö æ 1ö = 1 - ç1 - ÷ ç 1 - ÷ ç1 - ÷ è 2 ø è 3ø è 4ø 1 2 3 3 =1 - ´ ´ = 2 3 4 4 (d) The event follows binomial distribution with n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.; \ Variance = npq = 5/4. np = 4 ü 1 1 (b) npq = 2ý Þ q = 2 , p = 2 , n = 8 þ

æ 1 öæ 1 ö P ( X = 1) = 8C1 ç ÷ç ÷ è 2 øè 2 ø 1 1 1 = 8. 8 = 5 = 32 2 2

9 (d)

10 (d)

11 (b)

12 (b)

13 (d)

7

6.

15 (a)

(c) A and B will contradict each other if one speaks truth and other false . So , the required 4æ 3ö æ 4ö3 Probability = ç1 - ÷ + ç1 - ÷ 5è 4ø è 5ø4 4 1 1 3 7 ´ + ´ = 5 4 5 4 20 (b) P(E) = P ( 2 or 3 or 5 or 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 P ( F ) = P (1 or 2 or 3) = 0.15 + 0.23 + 0.12 = 0.50 P ( E Ç F ) = P(2 or 3) = 0.23 + 0.12 = 0.35 \ P( EUF ) = P( E ) + P ( F ) - P ( E Ç F ) = 0.62 + 0.50 - 0.35 = 0.77 (a) mean = np = 4 and variance = npq = 2 =

5.

14 (d)

\p=q=

1 and n = 8 2

M-191

Probability 6

7.

8.

2

æ1ö æ1ö \ P (2 success) = 8 C2 ç ÷ ç ÷ è2ø è2ø 28 28 = = 28 256 (b) For a particular house being selected 1 Probability = 3 P (all the persons apply for the same house) 1 æ1 1 1ö = ç ´ ´ ÷3 = . 9 è 3 3 3ø (c) According to Poission distribution, prob. of getting k successes is

P ( x = k ) = e -l

lk k!

P ( x ³ 2) = 1 - P ( x = 0) - P ( x = 1)

æ lö = 1 - e-l - e-l ç ÷ = 1 - 3 . è 1!ø e2 9.

(d)

P( X = r) =

e- m mr r!

P (at most 1 phone call) = P ( X £ 1) = P ( X = 0) + P( X = 1) 6 = e -5 + 5 ´ e -5 = e5 10. (d) Given : Probability of aeroplane I, scoring a target correctly i.e., P(I) = 0.3 probability of scoring a target correctly by aeroplane II, i.e. P(II) = 0.2 \ P( I ) = 1 – 0.3 = 0.7 \ The required probability

11.

= P ( I Ç II ) = P( I ).P( II ) =0.7× 0.2 = 0.14 (b) A pair of fair dice is thrown, the sample space S = (1, 1), (1, 2) (1, 3) .... = 36 Possibility of getting 9 are (5, 4), (4, 5), (6, 3), (3, 6) \ Possibilityof getting score 9 in a single throw =

4 1 = 36 9

\ Probability of getting score 9 exactly twice æ 1ö

2

æ

1ö

= 3C2 × ç ÷ . ç 1 - ÷ = 3! ´ 1 ´ 1 ´ 8 è 9ø è 9ø 2! 9 9 9 =

3.2! 1 1 8 8 ´ ´ ´ = 2! 9 9 9 243

1 12. (b) P(A) = 1/4, P(A/B) = , P(B/A) = 2/3 2 By conditional probability, P(A Ç B) = P(A) P(B/A) = P(B)P(A/B) 1 2 1 1 Þ ´ = P ( B ) ´ Þ P( B ) = 4 3 2 3 13. (d) We have 9 P (x ³ 1) ³ 10 9 Þ 1 – P (x = 0) ³ 10 0

n

Þ

9 æ1ö æ3ö 1 - nC0 ç ÷ ç ÷ ³ 10 è4ø è4ø

Þ

1-

9 æ 3ö ³ç ÷ 10 è 4 ø

n

n

æ 3ö æ 1ö çè ÷ø £ çè ÷ø 4 10 Taking log to the base 3/4, on both sides, we get æ3ö æ1ö n log3/4 ç ÷ ³ log3/4 ç ÷ 4 è ø è 10 ø - log10 10 Þ n ³ - log 3/4 10 = æ 3ö log10 ç ÷ è 4ø Þ

=

-1 log10 3 - log10 4

Þ n³

1 log10 4 - log10 3

14. (d) Let A º Sum of the digits is 8 B º Product of the digits is 0 Then A = {08, 17, 26, 35, 44} B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40,} A Ç B = { 08 } P ( A Ç B ) 150 1 = = \ P (A/B) = 14 P( B) 50 14 15. (a)

n( S ) = 9 C3

n( E ) = 3C1 ´ 4 C1 ´ 2 C1 Probability = 3 ´ 4 ´ 2 24 ´ 3! 24 ´ 6 2 = ´ 6! = = 9 9! 9 ´8´ 7 7 C3

EBD_7764 M-192

16.

17.

31 32 31 Þ 1 – p (no failure) ³ 32 31 5 Þ 1- p ³ 32 1 5 Þ p £ 32 1 Þ p£ 2 But p ³ 0

(b) p (at least one failure) ³

((

)

P Ac Ç B c Ç C

= = =

P (C )

) = P (( A È B) Ç C )

P ((1 - A È B) Ç C)

c

P (C )

P (C)

P ((1 - A - B + A Ç B ) Ç C ) P (C )

P(C) – P(A Ç C) – P(B Ç C) + P(A Ç B Ç C) P(C)

P(C) – P(A).P(C) – P(B)P(C) + 0 P(C) = 1 - P ( A) - P ( B) = P Ac - P ( B ) (b) Given sample space = {1,2,3,.....,8} Let Event A : Maximum of three numbers is 6. B : Minimum of three numbers is 3. This is the case of conditional probability We have to find P (minimum) is 3 when it is given that P (maximum) is 6. 2 C æ B ö P ( B Ç A) \ Pç ÷ = =5 1 P ( A) èAø C2 2 1 = = 10 5 =

( )

Mathematics (c) p = p (correct answer), q = p (wrong answer) 1 2 Þ p= , q = , n = 5 3 3 By using Binomial distribution Required probability 4

é 1ù Hence p lies in the interval ê0, ú . ë 2û (a) In this case æ C ö P(C Ç D) P (C ) = Pç ÷ = èDø P( D ) P ( D) Where, 0 £ P ( D) £ 1 , hence

æC ö P ç ÷ ³ P (C ) èDø 18. (d) P(Ac Ç Bc/C) =

19.

20.

5

æ 1ö 2 æ 1ö = 5 C4 ç ÷ · + 5C5 ç ÷ è 3ø 3 è 3ø 2 1 11 = 5· 5 + 5 = 5 3 3 3 21. (a) Given, 1 1 5 P( A È B ) = Þ P( A È B ) = 1 - = 6 6 6 1 1 3 P ( A) = Þ P ( A) = 1 - = 4 4 4 We know, P ( A È B ) = P( A) + P ( B ) - P ( A Ç B) 5 3 1 = + P( B ) Þ 6 4 4 1ö æ çèQ P( A Ç B) = ÷ø 4 1 Þ P( B) = 3 Q P ( A) ¹ P ( B ) so they are not equally likely. 3 1 1 ´ = 4 3 4 = P( A Ç B) So A & B are independent. 22. (b) We can apply binomial probability distribution We have n = 10 p = Probability of drawing a green ball = 15 3 = 25 5 3 2 Also q = 1 - = 5 5 Variance = npq 3 2 12 = 10 × ´ = 5 5 5

Also P ( A) ´ P( B ) =

23. (b)

Let A º {0, 1, 2, 3, 4, ......., 10} n (S) = 11C2 = 55 where 'S' denotes sample space Let E be the given event \ E º {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)} Þ n (E) = 6 n(E) 6 \ P(E) = = n(S) 55

M-193

Properties of Triangles

28

Properties of Triangles 1.

2.

3.

4.

The sides of a triangle are 3x+4y, 4x+3y and 5x+5y where x, y > 0 then the triangle is [2002] (a) right angled (b) obtuse angled (c) equilateral (d) none of these In a triangle with sides a, b, c, r1 > r2 > r3 (which are the ex-radii) then [2002] (a) a >b > c (b) a < b < c (c) a > b and b < c (d) a < b and b > c The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a , is [2003] (a)

a æp ö cot ç ÷ 4 è 2n ø

(b)

æp ö a cot ç ÷ ènø

(c)

a æp ö cot ç ÷ 2 è 2n ø

(d)

æp ö a cot ç ÷ è 2n ø

In a triangle ABC, medians AD and BE are drawn. p p If AD = 4, ÐDAB = and ÐABE = , then 6 3 the area of the D ABC is [2003] (a) (c)

5.

64 3 16 3

(b) (d)

8 3 32 3 3

2æ C ö 2 æ A ö 3b If in a DABC a cos ç ÷ + c cos ç ÷ = , then è2ø è2ø 2

the sides a, b and c

The sides of a triangle are sin a, cos a and 1 + sin a cos a for some 0 < a
0, c = 5x + 5y is the largest side \ C is the largest angle . Now cos C =

(3 x + 4 y )2 + (4 x + 3 y )3 - (5 x + 5 y ) 2 2(3x + 4 y)(4 x + 3 y)

-2 xy

> ; s-a s-b s-c Þ s – a < s – b < s – c Þ –a < –b < –c Þ a>b>c

2.

(a) r1 > r2 > r3 Þ

3.

(c)

a æ pö a æ pö tan ç ÷ = ; sin ç ÷ = è n ø 2r è n ø 2R r+R =

aé p pù cot + cos ec ú 2 êë n nû

M-195

Properties of Triangles

\ c is the greatest side and greatest angle is C a2 + b 2 - c 2 \ cos C = 2ab 2 2 1 sin a + cos a - 1 - sin a cos a == 2 2 sin a cos a \ C = 120°

O p n R

r N a

A

7.

B

p ù p é é ù cos + 1ú 2cos 2 ê aê a n 2n ú = ê ú= ê p p p ú 2ê ú 2 ê 2sin sin cos ú n û 2n 2n û ë ë =

4.

P

30º

E

3x =

P

60º

8.

B

D

C

2 8 4 AD = ; PD = ; 3 3 3 8/3 8 o tan 60 = or x = x 3 3 AP =

1 2

Area of DABD = ´ 4 ´

8 3 3

\ Area of DABC = 2 ´

5.

6.

=

16

60°

30° 40m

Q

R

x

From the figure y tan 60° = Þ y = 3 x ....................(1) x y x + 40 Þy= tan 30° = ...............(2) x + 40 3 From (1) and (2),

A

90º

T y

a p cot 2 2p

(d)

(d)

Let PB = x

x + 40

Þ x = 20m 3 (b) We know that for the circle circumscribing a right triangle, hypotenutse is the diameter As ÐC = 90o \ 2R = c Þ R =

16 3 3

=

32

3 3 3 3 [Q Median of a D divides it into two D¢s of equal area.] 2æ C ö 2 æ A ö 3b (b) If a cos ç ÷ + c cos ç ÷ = è 2ø è2ø 2 a[cos C + 1] + c[cos A + 1] = 3b (a + c ) + (a cos C + c cos B) = 3b a + c + b = 3b or a + c = 2b or a, b, c are in A.P. (c) Let a = sin a, b = cos a and c = 1 + sin a cos a Clearly a and b < 1 but c > 1 as sina > 0 and cosa > 0

c 2

1 ´ a ´b D 2 also r = = s a+b+c 2

Þr=

ab a+b+c

\ 2r + 2 R = =

2 ab +c a+b+c

2ab + ac + bc + c2 2ab + ac + bc + a 2b 2 = a+b+c a+b+c

(Q c

=

( a + b) 2 + ( a + b ) c a+b+c

2

= ( a + b)

= a 2 + b2

)

EBD_7764 M-196

9.

(b)

Mathematics

1 1 1 p1a = p2 b = p3b 2 2 2 p1, p2 , p3 , are in H.P..

D=

Þx=

2D 2D 2D are in H.P.. , , a b c 1 1 1 Þ , , are in H.P.. a b c Þ a, b, c are in A.P. Þ KsinA, K sin B, K sin C are in AP Þ sinA, sinB, sinC are in A.P. (a) In the D AOB, Ð AOB = 60°, and Ð OBA = Ð OAB (since OA = OB = AB radius of same circle). \ D AOB is a equilateral triangle. Let the height of tower is h

3 In D ABD h = tan45° =1 x+7

Þ

10.

h

h

Þ h=x+7Þh– Þ h=

7 3 3 –1

=7

3 +1 3 +1

(

)

7 3 3 +1 m 2 (b) If O is centre of polygon and AB is one of the side, then by figure

Þ h=

12.

´

3

O

C p/n

h

A

cos

60° 30°

O

B

r 1 1 3 = , , for n = 3, 4, 6 respectively.. R 2 2 2 (b) Given 3 sin P + 4cos Q = 6 ...(i) 4 sin Q + 3cos P = 1 ...(ii) Squaring and adding (i) & (ii) we get 9 sin2 P + 16cos2Q + 24 sin P cos Q + 16 sin2Q + 9cos2 P + 24 sin Q cos P = 36 + 1 = 37 Þ 9 (sin2P + cos2P) + 16 (sin2 Q + cos2 Q) + 24 (sinP cosQ + cosP sinQ) = 37 Þ 9 + 16 + 24 sin ( P + Q) = 37 [Q sin2q + cos2q = 1 and sin A cos B + cos A sin B = sin (A + B)] 1 Þ sin ( P + Q ) = 2 p 5p Þ P + Q = or 6 6 Þ R = 5 p or 5 p (Q P + Q + R = p) 6 6 5p p If R = then 0 < P , Q < 6 6

Þ

a

13.

B

11.

r

A

30° a

a

R

p r = n R

m. Given distance between two points A & B lie on boundary of circular park, subtends an angle of 60° at the foot of the tower is AB i.e. AB = a. A tower OC stands at the centre of a circular park. Angle of elevation of the top of the tower from A and B is 30°. h 1 h In DOAC, tan 30° = Þ = a 3 a a Þ h= 3 h (b) In D ABC = tan 60° = 3 x A

h

Þ cos Q < 1 and sin P < 11

60° B

x

45° C

7m

D

Þ 3 sin P + 4 cos Q < 2 p So, R = 6

1 2

M-197

Properties of Triangles

14.

(b) Letb + c = c + a = a + b = K in a triangleABC. 11

12

13

Þ b + c = 11 K, c + a = 12 K, a + b = 13 K On solving these, we get a = 7K, b = 6K, c = 5 K Now we know, cos A =

15.

b2 + c2 - a 2 2bc

A

O

B

...(i)

20 and from DBOD, tan 30° = ...(ii) x+ y From (i) and (ii), we have x = 20 and

3 Þ

20 x+ y 1 3

=

Þ tan a =

20 Þ 20 + y = 20 3 20 + y

So, y = 20( 3 - 1) i.e., speed = 20( 3 - 1) m/s

1 AB ) 2

1 4

B

A

20 x

...(1)

AC 1 AB 1 = = AP 2 AP 4

b a

D

y

From DAOC, tan 45° =

=

tan a =

AB 1 = AP 2

C

Let O be the point on the ground such that ÐAOC = 45° Let OC = x, Time t = 1 s

1

Let ÐAPC = a

20

20 C

Since AP = 2AB Þ

(Q C is the mid point) (\ AC =

36 K 2 + 25K 2 - 49 K 2 1 = = 2 ( 6 K )( 5K ) 5 (b) Let the speed be y m/sec. Let AC be the vertical pole of height 20 m.

45° 30° x

16. (d)

As tan (a + b) = Þ

P

tan a + tan b 1 - tan a tan b

tan a + tan b 1 - tan a tan b

AB é ù Qtan(a + b) = ê ú 1 AP ú = ê 1 2ê tan(a + b) = [ From(1) ]ú êë úû 2 1 + tan b 1 = Þ 4 1 2 1 - tan b 4

\ tan b =

2 9

EBD_7764